partTimeRunner Posted March 3, 2015 Share Posted March 3, 2015 I need to send the data gathered from a web form to my database. The connection works, and I can use the data gathered from post to make a query statement that works when used in mysql directly. My php script fails when I call the mysql_query() function. No error is displayed despite having the conditional right after it. My code: <?phpini_set ("display_errors", "1");error_reporting(E_ALL); $connection=mysqli_connect(/* not shown */); if (mysqli_connect_errno()) { echo "Failed to connect to database: " . mysqli_connect_error(); } $_clubName = $_POST['clubname']; $_clubRep = $_POST['clubrep']; $_meetingID = $_POST['meeting_id']; $_query = "UPDATE oh_checkin_2015 SET clubrep = '$_clubRep', meeting_2 = $_meetingID WHERE clubname = '$_clubName'"; $_result = mysql_query($_query); if (!$_result) die ("Database query failed: " . mysql _error()); else {echo 'successful query';}?> I have tried my best to locate the problem on my own and have looked up many other forums for a possible solution. If anyone can help me out with this, it would be greatly appreciated. Link to comment Share on other sites More sharing options...
justsomeguy Posted March 3, 2015 Share Posted March 3, 2015 You have a syntax error on the line with the if statement, you put a space after mysql. 1 Link to comment Share on other sites More sharing options...
partTimeRunner Posted March 5, 2015 Author Share Posted March 5, 2015 Ah! I need to be using VIM to spot these sort of things. Anyway, after fixing the conditional I did end up receiving errors: 'expects parameter 2 to be resource' and 'expects parameter 1 to be mysqli' which can be fixed by changing the mysql_ funtions into mysqli_ Problem solved. Thanks for the help. Link to comment Share on other sites More sharing options...
technologyparter Posted March 14, 2015 Share Posted March 14, 2015 See this example--- <?php// This could be supplied by a user, for example$firstname = 'fred';$lastname = 'fox';// Formulate Query// This is the best way to perform an SQL query// For more examples, see mysql_real_escape_string()$query = sprintf("SELECT firstname, lastname, address, age FROM friends WHERE firstname='%s' AND lastname='%s'", mysql_real_escape_string($firstname), mysql_real_escape_string($lastname));// Perform Query$result = mysql_query($query);// Check result// This shows the actual query sent to MySQL, and the error. Useful for debugging.if (!$result) { $message = 'Invalid query: ' . mysql_error() . "n"; $message .= 'Whole query: ' . $query; die($message);}// Use result// Attempting to print $result won't allow access to information in the resource// One of the mysql result functions must be used// See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc.while ($row = mysql_fetch_assoc($result)) { echo $row['firstname']; echo $row['lastname']; echo $row['address']; echo $row['age'];}// Free the resources associated with the result set// This is done automatically at the end of the scriptmysql_free_result($result);?> Link to comment Share on other sites More sharing options...
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