Php Mysql Error
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Mysql_query Error
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a community of 6.2 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 http://php.net/manual/en/function.mysql-error.php 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, I couldn't find it because the error is not specified on the web page. My question is, is there any way http://stackoverflow.com/questions/12227626/how-to-display-mysql-error-in-php 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 just use: $this->db_link->error to get the last error message. For all errors use $this->db_link->error_list. –hakre Sep 1 '12 at 12:31 add a comment| 7 Answers 7 active oldest votes up vote 17 down vote accepted Use this: mysqli_query($this->db_link, $query) or die(mysqli_error($this->db_link)); # mysqli_query($link,$q
here for a quick overview of the site Help Center Detailed answers to http://stackoverflow.com/questions/13909391/display-error-message-php-mysql any questions you might have Meta Discuss the workings and http://stackoverflow.com/questions/11623439/php-mysql-query-error 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 mysql error 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 2 i am unable to get the last 2 echos to work, even if the update php mysql error 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 dhere 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/MySQL query error? up vote 0 down vote favorite 1 Page is here: http://www.arkmediainc.com/ark20test/ark20test/portfolio.php $result = mysql_query( "SELECT * FROM foliobro WHERE itemid >= ".$startid." LIMIT 9" ); while($row = mysql_fetch_array($result)) { $itemid = $row['itemid']; $thumb_desc = $row['thumb_desc']; $title = $row['title']; $category = $row['category']; ... } I get this error code: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/arkmedia/public_html/ark20test/ark20test/portfolio.php on line 49 Line 49 is actually the while($row = mysql_fetch_array($result)) line. I'm trying to get the following 9 results, either (id's go from 1-11, with future expansion) 1-9, or 10-11+ What's wrong with my query, because I suspect that the query is not running properly. The query was taken straight from phpMyAdmin, so I don't see why it doesn't work. php mysql database share|improve this question edited May 19 '14 at 21:45 Tshepang 4,7061059103 asked Jul 24 '12 at 2:51 Tom Joyce 111 1 there's an error in your query. post a table row sample –maxhud Jul 24 '12 at 2:53 mysql_query() will also fail and return FALSE. a problem with the query, make sure the fields exist. –Novak Jul 24 '12 at 2:54 This has been asked so many times before. See the wiki here: stackoverflow.com/tags/php/info. Scroll down to the bottom of the page to get the answer. –Kemal Fadillah Jul 24 '12 at 2:54 1 Look into PDO. The mysql_* functions are deprecated as of PHP 5.3 –Bailey Parker Jul 24 '12 at 2:56 Is itemid an integer in the database? –Novak Jul 24 '12 at 3:03 add a comment| 3 Answers 3 active oldest votes up vote 3 down vote