englishcodemonkey Posted January 3, 2009 Share Posted January 3, 2009 Hi guys, So i'm trying to display an image with the code $picname = $row["fileContents"]; } echo "<img src = '$picname' >"; but if there are no results in internet explorer it shows a red X since it cant find the image, i've tried using "if(mysqli_num_rows($r)==0)" but this doesnt work. i'm using mysqli and so wondered if anyone had any other ideas.the query is shown below this: $q = "SELECT description from hosta where hosta_name='$hosta1'"; $r = mysqli_query ($dbc, $q); if ($r) { echo ' <table align="center" cellspacing="3" cellpadding="3" width="90%"> <tr> <td align="left"><b>Description</b></td> </tr>'; while ($row = mysqli_fetch_array($r,MYSQLI_ASSOC)) { echo ' <tr> <td align="justify">' .$row['description'] . '</td> </tr>'; } } echo '</table>'; $dbc = mysqli_connect ('****', '****', '****', '****'); $sqlimage = "SELECT * from image, hosta WHERE id=hosta_id and hosta_name = '$hosta'"; $sql_imgresult = mysqli_query($dbc,$sqlimage); while ( $row = mysqli_fetch_array ($sql_imgresult) ) { $picname = $row["fileContents"]; } echo "<img src = '$picname' >"; mysqli_close($dbc); ?> Link to comment https://forums.phpfreaks.com/topic/139376-if-statement-for-getting-no-results/ Share on other sites More sharing options...
fenway Posted January 4, 2009 Share Posted January 4, 2009 if() statements aren't mysql-related... prove that the query doesn't fail... check mysql error, make sure some rows are being returned, etc Link to comment https://forums.phpfreaks.com/topic/139376-if-statement-for-getting-no-results/#findComment-729261 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.