Php Echo Sql Error
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more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question mysql get last error 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 http://www.w3schools.com/php/func_mysqli_error.asp 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 query fails but I am having hard time looking for the reason behind this failure. Unfortunately, http://stackoverflow.com/questions/12227626/how-to-display-mysql-error-in-php 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 failed: %s\n", mysqli_connect_error()); exit; } $r = mysqli_query($this->db_link, $query); if($r==false) printf("error: %s\n", mysqli_errno($this->db_link)); php mysql share|improve this question asked Sep 1 '12 at 12:16 Traveling Salesman 56131640 You can ju
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 http://stackoverflow.com/questions/13909391/display-error-message-php-mysql 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 Display error message PHP Mysql up vote 1 down vote favorite sql error 2 i am unable to get the last 2 echos to work, even if the update query fails it still displays success. If anyone has any suggestions on this code to be improved on any line, please do! Success"; echo "
Your account password was successfully changed. Please click here to login.
"; } else { echo "Error
"; echo "Sorry, but a field was incorrect.
"; } } ?> Thanks in advance! php mysql share|improve this question edited Dec 17 '12 at 6:42 Sterling Archer 13.5k104173 asked Dec 17 '12 at 6:34 sparkones 26118 Please don't use mysql_query in new applications. It's terribly dangerous if not used perfectly which is an enormous nuisance to do, though I've seen you're at least trying here. You escaped two out of three variables and introduced a gigantic injection hole, though. Close enough is not good enough on the public internet. At the very least you should be using PDO unless you have a very good reason because when using SQL placeholders these mistakes are usually non-existent. –tadman Dec 17 '12 at 6:54 Thanks for this comment, i appreciate it and i will look into this. I haven't seen anything on it but ill