Mysql Error Checking In Php
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Mysqli_query Error
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Die(mysql_error())
Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a http://php.net/manual/en/mysqli.error.php minute: Sign up PHP MySQL error handling up vote 2 down vote favorite 1 Simple question (I hope). At the moment I am using the following code: mysql_query($sql) or header("Location: /error"); To prevent the rest of the script running I need to add exit; or die(). Is there any way to do that using the inline approach as above or do I http://stackoverflow.com/questions/4125559/php-mysql-error-handling need to use: $result = mysql_query($sql); if (!result) { header("Location: /error"); exit(); } Thanks php mysql share|improve this question edited Aug 28 '13 at 20:11 hdvianna 337211 asked Nov 8 '10 at 16:04 lethalMango 2,23583773 add a comment| 4 Answers 4 active oldest votes up vote 10 down vote accepted How about: function customDie($location) { header('Location: ' . $location); exit(); } mysql_query($sql) or customDie("/error"); share|improve this answer answered Nov 8 '10 at 16:07 Fosco 27.3k45593 Never thought about doing that, great idea. Thanks! –lethalMango Nov 8 '10 at 16:18 add a comment| up vote 5 down vote If you insist on doing things this way it is better to make a custom query method which handles all this. Something like function custom_mysql_query($query) { $doDebug=true; // Set to true when developing and false when you are deploying for real. $result=mysql_query($query); if(!$result) { if($doDebug) { // We are debugging so show some nice error output echo "Query failed\n
$query\n"; echo mysql_error(); // (Is that not the name) } else { // Might want an error message to the user here. } exit(); } } Then
here for a quick overview of the site Help Center Detailed answers to any questions you might http://stackoverflow.com/questions/12227626/how-to-display-mysql-error-in-php have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting http://www.php5-tutorial.com/mysql/handling-errors/ ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 mysql error million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up how to display MySql error in php up vote 14 down vote favorite In PHP, I am trying to execute a long MySql query that depends on the user input. However, my query fails with the following message, mysql error checking "Query Failed". Actually I have printed this message whenever the query fails but I am having hard time looking for the reason behind this failure. Unfortunately, I couldn't find it because the error is not specified on the web page. My question is, is there any way to display the error message that caused the failure on the web page. Thank you. Here's my code, $from= "Findings"; $where=""; if($service!= null) { $from = $from . ", ServiceType_Lookup"; $where= "Findings.ServiceType_ID= ServiceType_Lookup.ServiceType_ID AND ServiceType_Name= ". $service; if($keyword!= null) $where= $where . " AND "; } if( $keyword != null) { $where= $where . "Finding_ID LIKE '%$keyword%' OR ServiceType_ID LIKE '%$keyword%' OR Title LIKE '%$keyword%' OR RootCause_ID LIKE '%$keyword%' OR RiskRating_ID LIKE '%$keyword%' OR Impact_ID LIKE '%$keyword%' OR Efforts_ID LIKE '%$keyword%' OR Likelihood_ID LIKE '%$keyword%' OR Finding LIKE '%$keyword%' OR Implication LIKE '%$keyword%' OR Recommendation LIKE '%$keyword%' OR Report_ID LIKE '%$keyword%'"; } $query = "SELECT Finding_ID, ServiceType_ID, Title, RootCause_ID, RiskRating_ID, Impact_ID, Efforts_ID, Likelihood_ID, Finding, Implication, Recommendati
name or a keyword or something like that. By default, PHP will not show you exactly what the problem is, only that you wrote a query which is not entirely correct. Let's try writing a faulty query to see the response from PHP: $query = mysql_query("SELECT id, namme FROM test_users"); while($row = mysql_fetch_array($query)) echo $row['id'] . " - " . $row["name"] . " is from " . $row["country"] . "
"; This is the example we are using a lot in this part of the tutorial, but in this case, we have misspelled the name column to provoke an error, which we get: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in test.php on line 7 As you can see, the error is not thrown until we try using the resource returned by the mysql_query() function, which we do when we call the mysql_fetch_array() function, in my file located on line 7. The error is very generic and not very helpful. This is on purpose, because knowledge about your database structure makes your website more vulnerable to SQL injection attacks, a problem we will discuss later on. You might be able to spot the error and fix it in a lot of situations, but if not, you can use the mysql_error() function to get a bit more information abut the problem. This function simply returns any error returned from the last executed MySQL function. You should only use this function for finding and fixing problems, and then remove it again once the problem has been fixed. Here's the above example, but where we call the mysql_error() function to get more information: $query = mysql_query("SELECT id, namme FROM test_users"); while($row = mysql_fetch_array($query)) echo $row['id'] . " - " . $row["name"] . " is from " . $row["country"] . "
"; echo mysql_error(); This will give you a far more useful error message: Unknown column 'namme' in