Jqgrid dynamic filter with url data source












0















I have a Jqgrid as follows:



 jQuery("#jQGridDemo").jqGrid({
url: 'http://localhost:58404/JQGridHandler.ashx',
colNames: ['Property ID', 'Property Ref', 'Short Address', 'Scheme Code', 'Scheme Name', 'Property Type', 'Tenure Type', 'Status', 'Management Cost','Rent Charge Month','SC Charge Month'],
colModel: [
{ name: 'PropertyID', index: 'PropertyID', width: 70, align: "left", stype: 'text', sortable: true},
{name: 'PropertyType',width: 80},
{ name: 'TenureType', index: 'TenureType', width: 80, align: "center", sortable: true },
{ name: 'Status', index: 'Status', width: 75, align: "center", sortable: true },
],


The grid works and is populated with the Json returned from the URL.
However, I am trying to implement a dynamically populated drop-down-filter on the PropertyType column and have been looking at Oleg's answer here: How to add dynamic filter drop down using Jqgrid?



I have therefore added a "beforeProcessing" function:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data.rows, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
$(this).jqGrid("setColProp", 'PropertyType', {
stype: "select",
searchoptions: {
value: propertyValues
}
}).jqGrid('destroyFilterToolbar')
.jqGrid('filterToolbar', {
stringResult: true,
searchOnEnter: false,
defaultSearch: "cn"
});
},


My question is, how do I pass the data returned from the URL to the "beforeProcessing: function (data)" -



Any help would be appreciated.










share|improve this question























  • It is already passed. Not Sure that I understand. To this event is passed the data from the server before it is inserted into the grid. Also there are some differences in different jqGrid versions regarding this event - actually which version of jqGrid is used - Guriddo jqGrid, free-jqGrid or jqGrid <= 4.7?

    – Tony Tomov
    Nov 23 '18 at 20:26











  • I am using version 4.5.1. The debugger is telling me that rows is null. So I assume the data is not being passed to the function before it is inserted into the grid.

    – Sanxion
    Nov 26 '18 at 10:48













  • What is your datatype? If the datatype is local it will not be passed in version 4.5. For this version the data is passed only if the datatype is json or xml. The problem does not exist in Guriddo jqGrid versions.

    – Tony Tomov
    Nov 26 '18 at 11:55













  • My datatype is Json.

    – Sanxion
    Nov 26 '18 at 12:05











  • try to console.log the data. It is possible that data.rows property does not exist according to the jsonReader . just before the rows = data.rows. make a console.log of data only

    – Tony Tomov
    Nov 26 '18 at 12:39


















0















I have a Jqgrid as follows:



 jQuery("#jQGridDemo").jqGrid({
url: 'http://localhost:58404/JQGridHandler.ashx',
colNames: ['Property ID', 'Property Ref', 'Short Address', 'Scheme Code', 'Scheme Name', 'Property Type', 'Tenure Type', 'Status', 'Management Cost','Rent Charge Month','SC Charge Month'],
colModel: [
{ name: 'PropertyID', index: 'PropertyID', width: 70, align: "left", stype: 'text', sortable: true},
{name: 'PropertyType',width: 80},
{ name: 'TenureType', index: 'TenureType', width: 80, align: "center", sortable: true },
{ name: 'Status', index: 'Status', width: 75, align: "center", sortable: true },
],


The grid works and is populated with the Json returned from the URL.
However, I am trying to implement a dynamically populated drop-down-filter on the PropertyType column and have been looking at Oleg's answer here: How to add dynamic filter drop down using Jqgrid?



I have therefore added a "beforeProcessing" function:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data.rows, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
$(this).jqGrid("setColProp", 'PropertyType', {
stype: "select",
searchoptions: {
value: propertyValues
}
}).jqGrid('destroyFilterToolbar')
.jqGrid('filterToolbar', {
stringResult: true,
searchOnEnter: false,
defaultSearch: "cn"
});
},


My question is, how do I pass the data returned from the URL to the "beforeProcessing: function (data)" -



Any help would be appreciated.










share|improve this question























  • It is already passed. Not Sure that I understand. To this event is passed the data from the server before it is inserted into the grid. Also there are some differences in different jqGrid versions regarding this event - actually which version of jqGrid is used - Guriddo jqGrid, free-jqGrid or jqGrid <= 4.7?

    – Tony Tomov
    Nov 23 '18 at 20:26











  • I am using version 4.5.1. The debugger is telling me that rows is null. So I assume the data is not being passed to the function before it is inserted into the grid.

    – Sanxion
    Nov 26 '18 at 10:48













  • What is your datatype? If the datatype is local it will not be passed in version 4.5. For this version the data is passed only if the datatype is json or xml. The problem does not exist in Guriddo jqGrid versions.

    – Tony Tomov
    Nov 26 '18 at 11:55













  • My datatype is Json.

    – Sanxion
    Nov 26 '18 at 12:05











  • try to console.log the data. It is possible that data.rows property does not exist according to the jsonReader . just before the rows = data.rows. make a console.log of data only

    – Tony Tomov
    Nov 26 '18 at 12:39
















0












0








0








I have a Jqgrid as follows:



 jQuery("#jQGridDemo").jqGrid({
url: 'http://localhost:58404/JQGridHandler.ashx',
colNames: ['Property ID', 'Property Ref', 'Short Address', 'Scheme Code', 'Scheme Name', 'Property Type', 'Tenure Type', 'Status', 'Management Cost','Rent Charge Month','SC Charge Month'],
colModel: [
{ name: 'PropertyID', index: 'PropertyID', width: 70, align: "left", stype: 'text', sortable: true},
{name: 'PropertyType',width: 80},
{ name: 'TenureType', index: 'TenureType', width: 80, align: "center", sortable: true },
{ name: 'Status', index: 'Status', width: 75, align: "center", sortable: true },
],


The grid works and is populated with the Json returned from the URL.
However, I am trying to implement a dynamically populated drop-down-filter on the PropertyType column and have been looking at Oleg's answer here: How to add dynamic filter drop down using Jqgrid?



I have therefore added a "beforeProcessing" function:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data.rows, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
$(this).jqGrid("setColProp", 'PropertyType', {
stype: "select",
searchoptions: {
value: propertyValues
}
}).jqGrid('destroyFilterToolbar')
.jqGrid('filterToolbar', {
stringResult: true,
searchOnEnter: false,
defaultSearch: "cn"
});
},


My question is, how do I pass the data returned from the URL to the "beforeProcessing: function (data)" -



Any help would be appreciated.










share|improve this question














I have a Jqgrid as follows:



 jQuery("#jQGridDemo").jqGrid({
url: 'http://localhost:58404/JQGridHandler.ashx',
colNames: ['Property ID', 'Property Ref', 'Short Address', 'Scheme Code', 'Scheme Name', 'Property Type', 'Tenure Type', 'Status', 'Management Cost','Rent Charge Month','SC Charge Month'],
colModel: [
{ name: 'PropertyID', index: 'PropertyID', width: 70, align: "left", stype: 'text', sortable: true},
{name: 'PropertyType',width: 80},
{ name: 'TenureType', index: 'TenureType', width: 80, align: "center", sortable: true },
{ name: 'Status', index: 'Status', width: 75, align: "center", sortable: true },
],


The grid works and is populated with the Json returned from the URL.
However, I am trying to implement a dynamically populated drop-down-filter on the PropertyType column and have been looking at Oleg's answer here: How to add dynamic filter drop down using Jqgrid?



I have therefore added a "beforeProcessing" function:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data.rows, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
$(this).jqGrid("setColProp", 'PropertyType', {
stype: "select",
searchoptions: {
value: propertyValues
}
}).jqGrid('destroyFilterToolbar')
.jqGrid('filterToolbar', {
stringResult: true,
searchOnEnter: false,
defaultSearch: "cn"
});
},


My question is, how do I pass the data returned from the URL to the "beforeProcessing: function (data)" -



Any help would be appreciated.







filter drop-down-menu jqgrid






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 8:31









SanxionSanxion

54




54













  • It is already passed. Not Sure that I understand. To this event is passed the data from the server before it is inserted into the grid. Also there are some differences in different jqGrid versions regarding this event - actually which version of jqGrid is used - Guriddo jqGrid, free-jqGrid or jqGrid <= 4.7?

    – Tony Tomov
    Nov 23 '18 at 20:26











  • I am using version 4.5.1. The debugger is telling me that rows is null. So I assume the data is not being passed to the function before it is inserted into the grid.

    – Sanxion
    Nov 26 '18 at 10:48













  • What is your datatype? If the datatype is local it will not be passed in version 4.5. For this version the data is passed only if the datatype is json or xml. The problem does not exist in Guriddo jqGrid versions.

    – Tony Tomov
    Nov 26 '18 at 11:55













  • My datatype is Json.

    – Sanxion
    Nov 26 '18 at 12:05











  • try to console.log the data. It is possible that data.rows property does not exist according to the jsonReader . just before the rows = data.rows. make a console.log of data only

    – Tony Tomov
    Nov 26 '18 at 12:39





















  • It is already passed. Not Sure that I understand. To this event is passed the data from the server before it is inserted into the grid. Also there are some differences in different jqGrid versions regarding this event - actually which version of jqGrid is used - Guriddo jqGrid, free-jqGrid or jqGrid <= 4.7?

    – Tony Tomov
    Nov 23 '18 at 20:26











  • I am using version 4.5.1. The debugger is telling me that rows is null. So I assume the data is not being passed to the function before it is inserted into the grid.

    – Sanxion
    Nov 26 '18 at 10:48













  • What is your datatype? If the datatype is local it will not be passed in version 4.5. For this version the data is passed only if the datatype is json or xml. The problem does not exist in Guriddo jqGrid versions.

    – Tony Tomov
    Nov 26 '18 at 11:55













  • My datatype is Json.

    – Sanxion
    Nov 26 '18 at 12:05











  • try to console.log the data. It is possible that data.rows property does not exist according to the jsonReader . just before the rows = data.rows. make a console.log of data only

    – Tony Tomov
    Nov 26 '18 at 12:39



















It is already passed. Not Sure that I understand. To this event is passed the data from the server before it is inserted into the grid. Also there are some differences in different jqGrid versions regarding this event - actually which version of jqGrid is used - Guriddo jqGrid, free-jqGrid or jqGrid <= 4.7?

– Tony Tomov
Nov 23 '18 at 20:26





It is already passed. Not Sure that I understand. To this event is passed the data from the server before it is inserted into the grid. Also there are some differences in different jqGrid versions regarding this event - actually which version of jqGrid is used - Guriddo jqGrid, free-jqGrid or jqGrid <= 4.7?

– Tony Tomov
Nov 23 '18 at 20:26













I am using version 4.5.1. The debugger is telling me that rows is null. So I assume the data is not being passed to the function before it is inserted into the grid.

– Sanxion
Nov 26 '18 at 10:48







I am using version 4.5.1. The debugger is telling me that rows is null. So I assume the data is not being passed to the function before it is inserted into the grid.

– Sanxion
Nov 26 '18 at 10:48















What is your datatype? If the datatype is local it will not be passed in version 4.5. For this version the data is passed only if the datatype is json or xml. The problem does not exist in Guriddo jqGrid versions.

– Tony Tomov
Nov 26 '18 at 11:55







What is your datatype? If the datatype is local it will not be passed in version 4.5. For this version the data is passed only if the datatype is json or xml. The problem does not exist in Guriddo jqGrid versions.

– Tony Tomov
Nov 26 '18 at 11:55















My datatype is Json.

– Sanxion
Nov 26 '18 at 12:05





My datatype is Json.

– Sanxion
Nov 26 '18 at 12:05













try to console.log the data. It is possible that data.rows property does not exist according to the jsonReader . just before the rows = data.rows. make a console.log of data only

– Tony Tomov
Nov 26 '18 at 12:39







try to console.log the data. It is possible that data.rows property does not exist according to the jsonReader . just before the rows = data.rows. make a console.log of data only

– Tony Tomov
Nov 26 '18 at 12:39














1 Answer
1






active

oldest

votes


















0














In case your data does not have a rows property you can modifiy your initial code like this:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
....


The only that differ from your original code is replacing



rows = data.rows



with



rows = data,






share|improve this answer
























  • Thanks. That one change has made all the difference. I am new to the Jqgrid. Which version would you recommend I use?

    – Sanxion
    Nov 26 '18 at 14:45











  • It depends what you expect. You can use our commercial version Guriddo jqGrid with guaranteed support and development or you can use the open source fork free-jqGrid which has only support from Oleg without future Development

    – Tony Tomov
    Nov 26 '18 at 14:55












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0














In case your data does not have a rows property you can modifiy your initial code like this:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
....


The only that differ from your original code is replacing



rows = data.rows



with



rows = data,






share|improve this answer
























  • Thanks. That one change has made all the difference. I am new to the Jqgrid. Which version would you recommend I use?

    – Sanxion
    Nov 26 '18 at 14:45











  • It depends what you expect. You can use our commercial version Guriddo jqGrid with guaranteed support and development or you can use the open source fork free-jqGrid which has only support from Oleg without future Development

    – Tony Tomov
    Nov 26 '18 at 14:55
















0














In case your data does not have a rows property you can modifiy your initial code like this:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
....


The only that differ from your original code is replacing



rows = data.rows



with



rows = data,






share|improve this answer
























  • Thanks. That one change has made all the difference. I am new to the Jqgrid. Which version would you recommend I use?

    – Sanxion
    Nov 26 '18 at 14:45











  • It depends what you expect. You can use our commercial version Guriddo jqGrid with guaranteed support and development or you can use the open source fork free-jqGrid which has only support from Oleg without future Development

    – Tony Tomov
    Nov 26 '18 at 14:55














0












0








0







In case your data does not have a rows property you can modifiy your initial code like this:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
....


The only that differ from your original code is replacing



rows = data.rows



with



rows = data,






share|improve this answer













In case your data does not have a rows property you can modifiy your initial code like this:



 beforeProcessing: function (data) {
var propertyMap = {}, propertyValues = ":All", rows = data, i, symbol;
for (i = 0; i < rows.length; i++) {
symbol = rows[i].Symbol;
if (!propertyMap.hasOwnProperty(symbol)) {
propertyMap[symbol] = 1;
propertyValues += ";" + symbol + ":" + symbol;
}
}
....


The only that differ from your original code is replacing



rows = data.rows



with



rows = data,







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 26 '18 at 14:19









Tony TomovTony Tomov

1,6051710




1,6051710













  • Thanks. That one change has made all the difference. I am new to the Jqgrid. Which version would you recommend I use?

    – Sanxion
    Nov 26 '18 at 14:45











  • It depends what you expect. You can use our commercial version Guriddo jqGrid with guaranteed support and development or you can use the open source fork free-jqGrid which has only support from Oleg without future Development

    – Tony Tomov
    Nov 26 '18 at 14:55



















  • Thanks. That one change has made all the difference. I am new to the Jqgrid. Which version would you recommend I use?

    – Sanxion
    Nov 26 '18 at 14:45











  • It depends what you expect. You can use our commercial version Guriddo jqGrid with guaranteed support and development or you can use the open source fork free-jqGrid which has only support from Oleg without future Development

    – Tony Tomov
    Nov 26 '18 at 14:55

















Thanks. That one change has made all the difference. I am new to the Jqgrid. Which version would you recommend I use?

– Sanxion
Nov 26 '18 at 14:45





Thanks. That one change has made all the difference. I am new to the Jqgrid. Which version would you recommend I use?

– Sanxion
Nov 26 '18 at 14:45













It depends what you expect. You can use our commercial version Guriddo jqGrid with guaranteed support and development or you can use the open source fork free-jqGrid which has only support from Oleg without future Development

– Tony Tomov
Nov 26 '18 at 14:55





It depends what you expect. You can use our commercial version Guriddo jqGrid with guaranteed support and development or you can use the open source fork free-jqGrid which has only support from Oleg without future Development

– Tony Tomov
Nov 26 '18 at 14:55




















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