MySQLi query giving error when there no any result
I am trying to find and echo one result from my database. I am using it like below
<?php
error_reporting(E_ALL);
include_once("includes/connection.php");
$input = "OUT";
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
echo $answer;
?>
Its working fine when there some result but giving error like below when there no any result
Notice: Trying to get property 'answer' of non-object in C:xampphtdocsnewtest.php on line 5
I want handle this error using if else. I do not know much PHP and specially does not know more about query. Let me know if someone can help me for get out from this.
Thanks
php mysql sql mysqli
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
add a comment |
I am trying to find and echo one result from my database. I am using it like below
<?php
error_reporting(E_ALL);
include_once("includes/connection.php");
$input = "OUT";
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
echo $answer;
?>
Its working fine when there some result but giving error like below when there no any result
Notice: Trying to get property 'answer' of non-object in C:xampphtdocsnewtest.php on line 5
I want handle this error using if else. I do not know much PHP and specially does not know more about query. Let me know if someone can help me for get out from this.
Thanks
php mysql sql mysqli
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
add a comment |
I am trying to find and echo one result from my database. I am using it like below
<?php
error_reporting(E_ALL);
include_once("includes/connection.php");
$input = "OUT";
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
echo $answer;
?>
Its working fine when there some result but giving error like below when there no any result
Notice: Trying to get property 'answer' of non-object in C:xampphtdocsnewtest.php on line 5
I want handle this error using if else. I do not know much PHP and specially does not know more about query. Let me know if someone can help me for get out from this.
Thanks
php mysql sql mysqli
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
I am trying to find and echo one result from my database. I am using it like below
<?php
error_reporting(E_ALL);
include_once("includes/connection.php");
$input = "OUT";
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
echo $answer;
?>
Its working fine when there some result but giving error like below when there no any result
Notice: Trying to get property 'answer' of non-object in C:xampphtdocsnewtest.php on line 5
I want handle this error using if else. I do not know much PHP and specially does not know more about query. Let me know if someone can help me for get out from this.
Thanks
php mysql sql mysqli
php mysql sql mysqli
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
asked Nov 21 '18 at 7:04
Raju BhattRaju Bhatt
418
418
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
It's caused by the overuse of method chaining, why do you need to put every call in one long line?
You should check if an object is retrieved first...
$query = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1");
if ( $query && ($row = $query->fetch_object()) ) {
$answer = $row->answer;
echo $answer;
}
You should also look into prepared statements as this may lead to SQL injection and other problems. - How to create a secure mysql prepared statement in php?
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
Hi! This have solved my issue. Thanks
– Raju Bhatt
Nov 21 '18 at 7:16
Glad to help, if this has answered your question, please consider marking it as answered - meta.stackexchange.com/questions/5234/….
– Nigel Ren
Nov 21 '18 at 7:23
add a comment |
You can also use try{} catch {}
try {
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
}
catch (Exception $e) {
$errormsg = $e->getMessage();
}
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
add a comment |
Try this:
if ($answer) {
echo $answer;
} else {
die('No Answer Found');
}
Hope it will work.
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:12
StrangeStrange
807
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's caused by the overuse of method chaining, why do you need to put every call in one long line?
You should check if an object is retrieved first...
$query = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1");
if ( $query && ($row = $query->fetch_object()) ) {
$answer = $row->answer;
echo $answer;
}
You should also look into prepared statements as this may lead to SQL injection and other problems. - How to create a secure mysql prepared statement in php?
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
Hi! This have solved my issue. Thanks
– Raju Bhatt
Nov 21 '18 at 7:16
Glad to help, if this has answered your question, please consider marking it as answered - meta.stackexchange.com/questions/5234/….
– Nigel Ren
Nov 21 '18 at 7:23
add a comment |
It's caused by the overuse of method chaining, why do you need to put every call in one long line?
You should check if an object is retrieved first...
$query = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1");
if ( $query && ($row = $query->fetch_object()) ) {
$answer = $row->answer;
echo $answer;
}
You should also look into prepared statements as this may lead to SQL injection and other problems. - How to create a secure mysql prepared statement in php?
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
Hi! This have solved my issue. Thanks
– Raju Bhatt
Nov 21 '18 at 7:16
Glad to help, if this has answered your question, please consider marking it as answered - meta.stackexchange.com/questions/5234/….
– Nigel Ren
Nov 21 '18 at 7:23
add a comment |
It's caused by the overuse of method chaining, why do you need to put every call in one long line?
You should check if an object is retrieved first...
$query = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1");
if ( $query && ($row = $query->fetch_object()) ) {
$answer = $row->answer;
echo $answer;
}
You should also look into prepared statements as this may lead to SQL injection and other problems. - How to create a secure mysql prepared statement in php?
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
It's caused by the overuse of method chaining, why do you need to put every call in one long line?
You should check if an object is retrieved first...
$query = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1");
if ( $query && ($row = $query->fetch_object()) ) {
$answer = $row->answer;
echo $answer;
}
You should also look into prepared statements as this may lead to SQL injection and other problems. - How to create a secure mysql prepared statement in php?
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
edited Nov 21 '18 at 7:16
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
answered Nov 21 '18 at 7:13
Nigel RenNigel Ren
26.5k61833
26.5k61833
Hi! This have solved my issue. Thanks
– Raju Bhatt
Nov 21 '18 at 7:16
Glad to help, if this has answered your question, please consider marking it as answered - meta.stackexchange.com/questions/5234/….
– Nigel Ren
Nov 21 '18 at 7:23
add a comment |
Hi! This have solved my issue. Thanks
– Raju Bhatt
Nov 21 '18 at 7:16
Glad to help, if this has answered your question, please consider marking it as answered - meta.stackexchange.com/questions/5234/….
– Nigel Ren
Nov 21 '18 at 7:23
Hi! This have solved my issue. Thanks
– Raju Bhatt
Nov 21 '18 at 7:16
Hi! This have solved my issue. Thanks
– Raju Bhatt
Nov 21 '18 at 7:16
Glad to help, if this has answered your question, please consider marking it as answered - meta.stackexchange.com/questions/5234/….
– Nigel Ren
Nov 21 '18 at 7:23
Glad to help, if this has answered your question, please consider marking it as answered - meta.stackexchange.com/questions/5234/….
– Nigel Ren
Nov 21 '18 at 7:23
add a comment |
You can also use try{} catch {}
try {
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
}
catch (Exception $e) {
$errormsg = $e->getMessage();
}
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
add a comment |
You can also use try{} catch {}
try {
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
}
catch (Exception $e) {
$errormsg = $e->getMessage();
}
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
add a comment |
You can also use try{} catch {}
try {
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
}
catch (Exception $e) {
$errormsg = $e->getMessage();
}
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
You can also use try{} catch {}
try {
$answer = $mysqli->query("SELECT answer FROM faq WHERE question LIKE '%".$input."%' LIMIT 1")->fetch_object()->answer;
}
catch (Exception $e) {
$errormsg = $e->getMessage();
}
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
edited Nov 21 '18 at 7:38
Qirel
9,75262337
edited Nov 21 '18 at 7:38
Qirel
9,75262337
edited Nov 21 '18 at 7:38
Qirel
9,75262337
9,75262337
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
answered Nov 21 '18 at 7:18
KrishnaKrishna
404518
404518
add a comment |
add a comment |
Try this:
if ($answer) {
echo $answer;
} else {
die('No Answer Found');
}
Hope it will work.
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:12
StrangeStrange
807
add a comment |
Try this:
if ($answer) {
echo $answer;
} else {
die('No Answer Found');
}
Hope it will work.
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:12
StrangeStrange
807
add a comment |
Try this:
if ($answer) {
echo $answer;
} else {
die('No Answer Found');
}
Hope it will work.
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:12
StrangeStrange
807
Try this:
if ($answer) {
echo $answer;
} else {
die('No Answer Found');
}
Hope it will work.
edited Nov 21 '18 at 7:38
Qirel
9,75262337
answered Nov 21 '18 at 7:12
StrangeStrange
807
edited Nov 21 '18 at 7:38
Qirel
9,75262337
edited Nov 21 '18 at 7:38
Qirel
9,75262337
edited Nov 21 '18 at 7:38
Qirel
9,75262337
9,75262337
answered Nov 21 '18 at 7:12
StrangeStrange
807
answered Nov 21 '18 at 7:12
StrangeStrange
807
answered Nov 21 '18 at 7:12
StrangeStrange
807
807
add a comment |
add a comment |
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