[SOLVED] Error code 1050: Table 'registration' already exists

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  • delusion7
    New Member
    • Sep 2006
    • 37

    [SOLVED] Error code 1050: Table 'registration' already exists

    Trying to create a table and insert records from a webform and I keep getting this message:

    "Successful ly created the registration table.

    Unable to execute the query.

    Error code 1050: Table 'registration' already exists"

    Then after the table is created I re-run the query/webform and I don't receive the error message, but it doesn't update my table?
    When I go to mySQL it shows the table 'registration' but has no updated records.
    I'm 99.9% positive my code is correct, but I've also been working on this for the past 4hrs, so I could be a little out of it
    Code:
    <?php
    $DiverID = $_GET['diverID'];
    if (empty($DiverID))
    	exit("<p>You must enter a diver ID!  Click your browser's Back button to return to 	
    
    the previous page.</p>");
    
    $DBConnect = mysqli_connect("localhost", "root", "password")
    	Or die("Unable to connect to the database server.</p>"
    	."<p>Error code " . mysqli_connect_errno()
    	. ": " . mysqli_connect_error()) . "</p>";
    $DBName = "scuba_school";
    @mysqli_select_db($DBConnect, $DBName)
    	Or die("<p>Unable to select the database.</p>"
    	. "<p>Error code " . mysqli_errno($DBConnect)
    	. ": " . mysqli_error($DBConnect)) . "</p>";
    
    
    $TableName = "registration";
    $SQLstring = "SELECT * FROM $TableName";
    $QueryResult = @mysqli_query($DBConnect, $SQLstring);
    if (!$QueryResult) {
    	$SQLstring = "CREATE TABLE registration (diverID SMALLINT, class VARCHAR(40), days VARCHAR(40), time VARCHAR(40))";
    	$QueryResult = @mysqli_query($DBConnect, $SQLstring)
    		Or die("<p>Unable to create the registration table.</p>"
    		. "<p>Error code " . mysqli_errno($DBConnect)
    		. ": " .mysqli_error($DBConnect)) . "</p>";
    	echo "<p>Successfully created the registration table.</p>";
    		   }
    
    $Class = $_GET['class'];
    $Days = $_GET['days'];
    $Time = $_GET['time'];
    $SQlstring = "INSERT INTO $TableName VALUES('$DiverID', '$Class', '$Days', '$Time')";
    $QueryResult = @mysqli_query($DBConnect, $SQLstring)
    	Or die("<p>Unable to execute the query.</p>"
    	. "<p>Error code " . mysqli_errno($DBConnect)
    	. ": " .mysqli_error($DBConnect)) . "</p>";
    
    mysqli_close($DBConnect);
  • delusion7
    New Member
    • Sep 2006
    • 37

    #2
    Ok, well sometimes i guess a little sleep is needed to solve a problem. My variables were not the same in two sections of my script. User error, please ignore.. : )

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