Sum values from a column in database [closed]












-1














Good morning,



I 've been struggling with the simple task of creating a sum of numerical values from the database. What I want to do is to create a loop within another loop.



The first loop is fetching 'project' from database and report the values in a table on my page and the bracket closes after the last column of the page. Now the part that i want to do is to use the $id to fetch the values from 'budget' in the database, and the ones that have the project_id='$id' to sum and echo them in another cell of the same table.



My code so far is the following:



<?php   
include 'dbcon.php';

$query1=mysqli_query($con,"select * from project ORDER BY project_id ASC")or die(mysqli_error($con));
while ($row1=mysqli_fetch_array($query1)){
$id=$row['project_id'];
$total=0;
$query = mysqli_query($con,"SELECT * FROM budget WHERE project_id='$id'");
while ($row=mysqli_fetch_array($query)){
$total=$total+$row['amount'];}
?>
<tr>
<td colspan="2"><?php echo $row1['project_name'];?></td>
<td><?php echo $total;?></td>
<td><?php echo $row1['project_status'];?></td>
</tr>
<?php }?>


Now what I am getting is 0 in the table and not the summary. I have also tryid some solution I found in the site like:



$query = mysqli_query($con,"SELECT sum(amount) as total FROM budget WHERE project_id='$id'");


No luck with that one either. Now i have two explanations. Either the loop within a loop is not build correctly by me (i believe that setting $id=$row1['project_id'] will push the project_id to the next loop and will not change until the second loop finishes) or i missing something really silly here...










share|improve this question















closed as off-topic by Rob Nov 30 '18 at 3:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Rob

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    You know MySQL has a SUM() function, yes?
    – CD001
    Nov 20 '18 at 9:04










  • Hello CD001, I have tryed it like: '$total = new PDO; foreach($total->query('SELECT SUM(amount) FROM budget where project_id='$id'') as $row)' without success. I think i am missing something really simple here..
    – juamx214
    Nov 20 '18 at 9:13












  • @juamx214 why are you using two queries ? In your current code, you will get only the TOTAL of last project that you get from query one.
    – Sajjad Ali
    Nov 20 '18 at 9:59










  • If your aim is to get the total of a specific project then you should have to get SUM of only that particular subject (by getting the id of subject) rather than getting all the projects, which is surely not a good practice.
    – Sajjad Ali
    Nov 20 '18 at 10:00










  • I question the edit where they attempted to answer. I also question it being approved. I rolled it back.
    – Funk Forty Niner
    Nov 20 '18 at 13:58


















-1














Good morning,



I 've been struggling with the simple task of creating a sum of numerical values from the database. What I want to do is to create a loop within another loop.



The first loop is fetching 'project' from database and report the values in a table on my page and the bracket closes after the last column of the page. Now the part that i want to do is to use the $id to fetch the values from 'budget' in the database, and the ones that have the project_id='$id' to sum and echo them in another cell of the same table.



My code so far is the following:



<?php   
include 'dbcon.php';

$query1=mysqli_query($con,"select * from project ORDER BY project_id ASC")or die(mysqli_error($con));
while ($row1=mysqli_fetch_array($query1)){
$id=$row['project_id'];
$total=0;
$query = mysqli_query($con,"SELECT * FROM budget WHERE project_id='$id'");
while ($row=mysqli_fetch_array($query)){
$total=$total+$row['amount'];}
?>
<tr>
<td colspan="2"><?php echo $row1['project_name'];?></td>
<td><?php echo $total;?></td>
<td><?php echo $row1['project_status'];?></td>
</tr>
<?php }?>


Now what I am getting is 0 in the table and not the summary. I have also tryid some solution I found in the site like:



$query = mysqli_query($con,"SELECT sum(amount) as total FROM budget WHERE project_id='$id'");


No luck with that one either. Now i have two explanations. Either the loop within a loop is not build correctly by me (i believe that setting $id=$row1['project_id'] will push the project_id to the next loop and will not change until the second loop finishes) or i missing something really silly here...










share|improve this question















closed as off-topic by Rob Nov 30 '18 at 3:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Rob

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    You know MySQL has a SUM() function, yes?
    – CD001
    Nov 20 '18 at 9:04










  • Hello CD001, I have tryed it like: '$total = new PDO; foreach($total->query('SELECT SUM(amount) FROM budget where project_id='$id'') as $row)' without success. I think i am missing something really simple here..
    – juamx214
    Nov 20 '18 at 9:13












  • @juamx214 why are you using two queries ? In your current code, you will get only the TOTAL of last project that you get from query one.
    – Sajjad Ali
    Nov 20 '18 at 9:59










  • If your aim is to get the total of a specific project then you should have to get SUM of only that particular subject (by getting the id of subject) rather than getting all the projects, which is surely not a good practice.
    – Sajjad Ali
    Nov 20 '18 at 10:00










  • I question the edit where they attempted to answer. I also question it being approved. I rolled it back.
    – Funk Forty Niner
    Nov 20 '18 at 13:58
















-1












-1








-1







Good morning,



I 've been struggling with the simple task of creating a sum of numerical values from the database. What I want to do is to create a loop within another loop.



The first loop is fetching 'project' from database and report the values in a table on my page and the bracket closes after the last column of the page. Now the part that i want to do is to use the $id to fetch the values from 'budget' in the database, and the ones that have the project_id='$id' to sum and echo them in another cell of the same table.



My code so far is the following:



<?php   
include 'dbcon.php';

$query1=mysqli_query($con,"select * from project ORDER BY project_id ASC")or die(mysqli_error($con));
while ($row1=mysqli_fetch_array($query1)){
$id=$row['project_id'];
$total=0;
$query = mysqli_query($con,"SELECT * FROM budget WHERE project_id='$id'");
while ($row=mysqli_fetch_array($query)){
$total=$total+$row['amount'];}
?>
<tr>
<td colspan="2"><?php echo $row1['project_name'];?></td>
<td><?php echo $total;?></td>
<td><?php echo $row1['project_status'];?></td>
</tr>
<?php }?>


Now what I am getting is 0 in the table and not the summary. I have also tryid some solution I found in the site like:



$query = mysqli_query($con,"SELECT sum(amount) as total FROM budget WHERE project_id='$id'");


No luck with that one either. Now i have two explanations. Either the loop within a loop is not build correctly by me (i believe that setting $id=$row1['project_id'] will push the project_id to the next loop and will not change until the second loop finishes) or i missing something really silly here...










share|improve this question















Good morning,



I 've been struggling with the simple task of creating a sum of numerical values from the database. What I want to do is to create a loop within another loop.



The first loop is fetching 'project' from database and report the values in a table on my page and the bracket closes after the last column of the page. Now the part that i want to do is to use the $id to fetch the values from 'budget' in the database, and the ones that have the project_id='$id' to sum and echo them in another cell of the same table.



My code so far is the following:



<?php   
include 'dbcon.php';

$query1=mysqli_query($con,"select * from project ORDER BY project_id ASC")or die(mysqli_error($con));
while ($row1=mysqli_fetch_array($query1)){
$id=$row['project_id'];
$total=0;
$query = mysqli_query($con,"SELECT * FROM budget WHERE project_id='$id'");
while ($row=mysqli_fetch_array($query)){
$total=$total+$row['amount'];}
?>
<tr>
<td colspan="2"><?php echo $row1['project_name'];?></td>
<td><?php echo $total;?></td>
<td><?php echo $row1['project_status'];?></td>
</tr>
<?php }?>


Now what I am getting is 0 in the table and not the summary. I have also tryid some solution I found in the site like:



$query = mysqli_query($con,"SELECT sum(amount) as total FROM budget WHERE project_id='$id'");


No luck with that one either. Now i have two explanations. Either the loop within a loop is not build correctly by me (i believe that setting $id=$row1['project_id'] will push the project_id to the next loop and will not change until the second loop finishes) or i missing something really silly here...







php loops sum






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 30 '18 at 3:25









Rob

23.6k115274




23.6k115274










asked Nov 20 '18 at 9:02









juamx214

24




24




closed as off-topic by Rob Nov 30 '18 at 3:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Rob

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Rob Nov 30 '18 at 3:25


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question was caused by a problem that can no longer be reproduced or a simple typographical error. While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. This can often be avoided by identifying and closely inspecting the shortest program necessary to reproduce the problem before posting." – Rob

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    You know MySQL has a SUM() function, yes?
    – CD001
    Nov 20 '18 at 9:04










  • Hello CD001, I have tryed it like: '$total = new PDO; foreach($total->query('SELECT SUM(amount) FROM budget where project_id='$id'') as $row)' without success. I think i am missing something really simple here..
    – juamx214
    Nov 20 '18 at 9:13












  • @juamx214 why are you using two queries ? In your current code, you will get only the TOTAL of last project that you get from query one.
    – Sajjad Ali
    Nov 20 '18 at 9:59










  • If your aim is to get the total of a specific project then you should have to get SUM of only that particular subject (by getting the id of subject) rather than getting all the projects, which is surely not a good practice.
    – Sajjad Ali
    Nov 20 '18 at 10:00










  • I question the edit where they attempted to answer. I also question it being approved. I rolled it back.
    – Funk Forty Niner
    Nov 20 '18 at 13:58
















  • 1




    You know MySQL has a SUM() function, yes?
    – CD001
    Nov 20 '18 at 9:04










  • Hello CD001, I have tryed it like: '$total = new PDO; foreach($total->query('SELECT SUM(amount) FROM budget where project_id='$id'') as $row)' without success. I think i am missing something really simple here..
    – juamx214
    Nov 20 '18 at 9:13












  • @juamx214 why are you using two queries ? In your current code, you will get only the TOTAL of last project that you get from query one.
    – Sajjad Ali
    Nov 20 '18 at 9:59










  • If your aim is to get the total of a specific project then you should have to get SUM of only that particular subject (by getting the id of subject) rather than getting all the projects, which is surely not a good practice.
    – Sajjad Ali
    Nov 20 '18 at 10:00










  • I question the edit where they attempted to answer. I also question it being approved. I rolled it back.
    – Funk Forty Niner
    Nov 20 '18 at 13:58










1




1




You know MySQL has a SUM() function, yes?
– CD001
Nov 20 '18 at 9:04




You know MySQL has a SUM() function, yes?
– CD001
Nov 20 '18 at 9:04












Hello CD001, I have tryed it like: '$total = new PDO; foreach($total->query('SELECT SUM(amount) FROM budget where project_id='$id'') as $row)' without success. I think i am missing something really simple here..
– juamx214
Nov 20 '18 at 9:13






Hello CD001, I have tryed it like: '$total = new PDO; foreach($total->query('SELECT SUM(amount) FROM budget where project_id='$id'') as $row)' without success. I think i am missing something really simple here..
– juamx214
Nov 20 '18 at 9:13














@juamx214 why are you using two queries ? In your current code, you will get only the TOTAL of last project that you get from query one.
– Sajjad Ali
Nov 20 '18 at 9:59




@juamx214 why are you using two queries ? In your current code, you will get only the TOTAL of last project that you get from query one.
– Sajjad Ali
Nov 20 '18 at 9:59












If your aim is to get the total of a specific project then you should have to get SUM of only that particular subject (by getting the id of subject) rather than getting all the projects, which is surely not a good practice.
– Sajjad Ali
Nov 20 '18 at 10:00




If your aim is to get the total of a specific project then you should have to get SUM of only that particular subject (by getting the id of subject) rather than getting all the projects, which is surely not a good practice.
– Sajjad Ali
Nov 20 '18 at 10:00












I question the edit where they attempted to answer. I also question it being approved. I rolled it back.
– Funk Forty Niner
Nov 20 '18 at 13:58






I question the edit where they attempted to answer. I also question it being approved. I rolled it back.
– Funk Forty Niner
Nov 20 '18 at 13:58














1 Answer
1






active

oldest

votes


















-1














You might need to change this. I think you are doing something wrong here:



 $id=$row['project_id'];


To this:



 $id=$row1['project_id'];


See edit code






share|improve this answer






























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    -1














    You might need to change this. I think you are doing something wrong here:



     $id=$row['project_id'];


    To this:



     $id=$row1['project_id'];


    See edit code






    share|improve this answer




























      -1














      You might need to change this. I think you are doing something wrong here:



       $id=$row['project_id'];


      To this:



       $id=$row1['project_id'];


      See edit code






      share|improve this answer


























        -1












        -1








        -1






        You might need to change this. I think you are doing something wrong here:



         $id=$row['project_id'];


        To this:



         $id=$row1['project_id'];


        See edit code






        share|improve this answer














        You might need to change this. I think you are doing something wrong here:



         $id=$row['project_id'];


        To this:



         $id=$row1['project_id'];


        See edit code







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 21 '18 at 13:52









        Funk Forty Niner

        80.5k1247101




        80.5k1247101










        answered Nov 20 '18 at 9:18









        Akhtar Munir

        6811




        6811















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