baber_abbasi Posted May 26, 2003 Share Posted May 26, 2003 I am using a script to display image from database. I want to display all images from my table onto the page. The code I am using is following and I will appreciate ur help if necessary modifications are made into it as required. \" $dbServer = \"localhost\"; $dbDatabase = \"blob\"; $dbUser = \"root\"; $dbPass = \"\"; $sConn = mysql_connect($dbServer, $dbUser, $dbPass) or die(\"Couldn\'t connect to database server\"); $dConn = mysql_select_db($dbDatabase, $sConn) or die(\"Couldn\'t connect to database $dbDatabase\"); $result = mysql_query(\"SELECT * FROM myBlobs\"); if(mysql_num_rows($result) > 0) { $row = mysql_fetch_array($result); do{ $dd = $row[\'blobData\']; print($dd.\"<BR>\"); $row = mysql_fetch_array($result); }while($row); }else{echo\"No record\";} \" From above code, it displays only first image and don\'t increment to next which it should be. Please tell me how sould I resolve this problem. Thankx Quote Link to comment https://forums.phpfreaks.com/topic/509-image-displaying-from-database/ Share on other sites More sharing options...
shivabharat Posted May 26, 2003 Share Posted May 26, 2003 Try this :roll: $result = mysql_query("SELECT * FROM myBlobs"); if(mysql_num_rows($result) > \'0\') { do{ $dd = $row[\'blobData\']; print($dd."<BR>"); $row = mysql_fetch_array($result); }while($row=mysql_fetch_array($result)); }else{echo"No record";} Quote Link to comment https://forums.phpfreaks.com/topic/509-image-displaying-from-database/#findComment-1701 Share on other sites More sharing options...
baber_abbasi Posted May 26, 2003 Author Share Posted May 26, 2003 Thankx for ur help. I tried this but get same output thatis only forst image from table is displayed. Couldu plz check it again? Thankx again. Quote Link to comment https://forums.phpfreaks.com/topic/509-image-displaying-from-database/#findComment-1707 Share on other sites More sharing options...
shivabharat Posted May 27, 2003 Share Posted May 27, 2003 Hope thsi works :roll: $result = mysql_query("SELECT * FROM myBlobs"); if(mysql_num_rows($result) > \'0\') { do{ $dd = $row[\'blobData\']; print($dd."<BR>"); }while($row=mysql_fetch_array($result)); }else{echo"No record";} Quote Link to comment https://forums.phpfreaks.com/topic/509-image-displaying-from-database/#findComment-1720 Share on other sites More sharing options...
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