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POMysql Error when trying to spit out database elements
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  1. POMysql Error when trying to spit out database elements
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    copied!<p>I keep getting this error :</p> <pre><code>Warning: mysqli_error() expects exactly 1 parameter, 0 given in /Applications/MAMP/htdocs/dbViewer.php on line 71 </code></pre> <p>Could not get data: </p> <p>from this code</p> <pre><code>$sql = "SELECT DISTINCT TABLE_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE COLUMN_NAME IN ('Version') AND TABLE_SCHEMA = 'wp_plugin_db'"; $result = mysqli_query($con,$sql); if(! $result ) { die('Could not get data: ' . mysqli_error()); } $arrayCount = 0; while ($row=mysqli_fetch_array($result)) { $tableNames[$arrayCount] = $row[0]; $arrayCount++; } foreach ($tableNames as $siteName) { $siteName = mysqli_real_escape_string($con,$siteName); $sql="SELECT Plugin_Name, Version, WPVersion FROM `".$siteName."` ORDER BY Plugin_Name"; $result=mysqli_query($con,$sql); if(! $result ) { die('Could not get data: ' . mysqli_error()); } echo "Website Name: $siteName ---- " ; while($row=mysqli_fetch_array($result,MYSQLI_ASSOC)) { echo " Plugin Name :{$row['Plugin_Name']} ". " Version : {$row['Version']} ". " Wordpress Version : {$row['WPVersion']} ". " | "; } echo "&lt;br&gt;"; } mysqli_close($con); </code></pre> <p>The error says <code>line 71</code> which is <code>die('Could not get data: ' . mysqli_error());</code></p> <p>The real error is happening at <code>$sql="SELECT Plugin_Name, WPVersion, Version FROM ".$siteName." ORDER BY Plugin_Name";</code> </p> <p>I know this because when I take out WPVersion from the selec and comment out the echo for that element it works without error. Please help I can't seem to figure this out, Is it syntax? I feel so stupid haha.</p> <p>Thank you in advance!</p>
 

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