Php Error Mysql
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Mysqli_query Error
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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 how http://php.net/manual/en/function.mysql-error.php 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 just
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 http://stackoverflow.com/questions/11623439/php-mysql-query-error the company Business Learn more about hiring developers or posting ads with us Stack Overflow http://www.plus2net.com/php_tutorial/php_mysql_query.php 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 $result = mysql_query( "SELECT mysql error * FROM foliobro WHERE itemid >= ".$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) php error mysql 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 Change your code to: $result = mysql_query("SELECT * FROM foliobro WHERE itemid >= ".$startid." LIMIT 9"); if (!$result) { die(mysql_error()); } Your query is failing and you are passing an
& Database Submit Sign UP Login × Login Username Password Login Cancel Not a member? Sign Up Forgot Password? Contact Us PHP MySQL query with error printing How to write SQL using PHP to handle the data in MySQL database? In any database driven script we have to update, add, modify, data in the tables. By using PHP we can do all this using different functions available in PHP. We will start with very basic function, which will execute any query written in sql and can be applied to MySQL database. SQL Structured Query Language or popularly known as SQL is an universal language to handle database. An introduction and different types of sql command like select, insert, update etc you will get in the sql section of this site. There are some advance SQL commands like left join, linking of tables etc to study. If you are not comfortable with SQL any time you can refer the materials in sql section. There are three steps invoved in this process. Connection to database Build the query and execute Display the data First ensure that you have established your mysql connection through PHP. To get the full details on php mysql connection you can read the article here. If you are using PDO then start with PDO connection string here. PHP Functions & SQL Let us start with the function required to execute one query in PHP. Once you have connection established then we can execute sql command by using PHP function mysql_query(). Here is the syntax of the function. Let us first write the query and store in a variable. We will write a query to create table.$query="CREATE TABLE student ( id int(2) NOT NULL auto_increment, name varchar(50) NOT NULL default '', class varchar(10) NOT NULL default '', mark int(3) NOT NULL default '0', PRIMARY KEY (id) ) TYPE=MyISAM"; We have stored the sql create query in a variable $query and we will pass this as a parameter to the function like below. $rt=mysql_query($query); The above command will execute the query ( stored in variable $query) and we can check the status of the query ( successful or not ) by checking the status of $rt. $rt will be true of the query is successfully executed or it will return false. We will use php if command to check the status of the query. if($rt){echo " Command is successful ";} else {echo "Command is not successful ";} So from the above line we can know that the query has worked or failed. But we will not come to know what the error is if the query has failed. To get the error message we have to use an