Php Sql Error Checking
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here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss http://stackoverflow.com/questions/12227626/how-to-display-mysql-error-in-php the workings and policies of this site About Us Learn more about http://stackoverflow.com/questions/1918624/php-try-and-catch-for-sql-insert Stack Overflow the company Business Learn more about hiring developers or posting 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 million programmers, just like you, helping each sql error 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, "Query Failed". Actually I have printed this message whenever the php sql error 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%'"; } $query = "SELECT Finding_ID, ServiceType_ID, Title, RootCause_ID, RiskRating_ID, Impact_ID, Efforts_ID, Likelihood_ID, Finding, Implication, Recommendation, Report_ID FROM ".$from . " WHERE " . $where; echo "wala 2eshiq"; $this->result = $this->db_link->query($query); if (!$this->result) { printf("Query faile
here for a quick overview of the site Help Center Detailed answers to any questions you might 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 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 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up PHP Try and Catch for SQL Insert up vote 18 down vote favorite 1 I have a page on my website (high traffic) that does an insert on every page load. I am curious of the fastest and safest way to (catch an error) and continue if the system is not able to do the insert into MySQL. Should I use try/catch or die or something else. I want to make sure the insert happens but if for some reason it can't I want the page to continue to load anyway. ... $db = mysql_select_db('mobile', $conn); mysql_query("INSERT INTO redirects SET ua_string = '$ua_string'") or die('Error #10'); mysql_close($conn); ... php mysql performance error-handling share|improve this question asked Dec 16 '09 at 23:51 meme 5,64621218 On the use of "or die": phpfreaks.com/blog/or-die-must-die –outis Dec 17 '09 at 0:55 1 As for exceptions vs checking return values, it depends on how many points might generate errors. With one or two points, I'd go with error checking, as it's more performant and just as readable in this case. Once you hit three or more error checks in a code block, exceptions become more readable. It's all about reducing cyclomatic complexity. Note that this covers the point you handle the error; if you're talking about signaling errors, you'll wind up with different guidelines. –outis Dec 17 '09 at 0:59 add a comment| 8 Answers 8 active oldest votes up vote 23 down vote Checking the documentation shows that its returns false on an error. So use the return status rather than or die(). It will return false if it fails, which y