smitagodbole Posted August 27, 2010 Share Posted August 27, 2010 Hi, I am a newbie to this forum and php as well. I have a script which inserts data and image file into mysql database. Now I want to display this image on web page but unable to do it. I tried a lot but couldn't find out any solution. I would appreciate your help. <?php $con = mysql_connect("localhost", "root", "xx"); if(!$con) { die('Could not connect:' .mysql_error()); } mysql_select_db("yyy",$con); $memberid = '$_POST[memberid]'; $fname = '$_POST[fname]'; $lname = '$_POST[lname]'; $gender= '$_POST[gender]'; $add1 = '$_POST[add1]'; $add2 = '$_POST[add2]'; $city = '$_POST[city]'; $state = '$_POST[state]'; $zip = '$_POST[zip]'; $country = '$_POST[country]'; $photo1 = '$_FILES[image][name]'; echo "$photo1"; //if ($memberid= $_GET['memberid']){ $query="select memberid,fname,lname,gender,add1,add2,city,state,zip,country,photo1 from vadhuvar_member where memberid =(select max(memberid)from vadhuvar_member)"; $result=mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($result)) { $memberid = $row['memberid']; $fname = $row['fname']; $lname = $row['lname']; $gender = $row['gender']; $add1 = $row['add1']; $add2 = $row['add2']; $city = $row['city']; $state = $row['state']; $zip = $row['zip']; $country = $row['country']; //$photo1=chunk_split(base64_encode($photo1)); $photo1 = $row['photo1']; //echo base64_encode($photo1); echo "Name:$fname $lname<br>"; //echo "Name:$_POST[fname]$_POST[lname]<br>"; echo "Gender:$gender<br>"; echo "Address:$add1 $add2<br>"; echo "$city $state $zip<br>"; echo "$country<br>"; //echo "<img src=".upload/$photo1 ." alt="" width="200" height="300" /><br />"; } //} ?> <!--img src="<?php echo '.upload/$_FILES[image][name]?memberid=$row[memberid]'?>"alt="" width="200" height="300"/-->; <?php echo "<img src=". upload/$row[photo1]" alt="" width="200" height="300">";?> <!--img src="<?php echo ".\upload\$photo1" ?>" alt="" width="200" height="300"/-->; <? set_magic_quotes_runtime(1); // turn back on ?> Thanks Smita Quote Link to comment Share on other sites More sharing options...
Vince889 Posted August 27, 2010 Share Posted August 27, 2010 Hi, I am a newbie to this forum and php as well. I have a script which inserts data and image file into mysql database. Now I want to display this image on web page but unable to do it. I tried a lot but couldn't find out any solution. I would appreciate your help. <?php $con = mysql_connect("localhost", "root", "xx"); if(!$con) { die('Could not connect:' .mysql_error()); } mysql_select_db("yyy",$con); $memberid = '$_POST[memberid]'; $fname = '$_POST[fname]'; $lname = '$_POST[lname]'; $gender= '$_POST[gender]'; $add1 = '$_POST[add1]'; $add2 = '$_POST[add2]'; $city = '$_POST[city]'; $state = '$_POST[state]'; $zip = '$_POST[zip]'; $country = '$_POST[country]'; $photo1 = '$_FILES[image][name]'; echo "$photo1"; //if ($memberid= $_GET['memberid']){ $query="select memberid,fname,lname,gender,add1,add2,city,state,zip,country,photo1 from vadhuvar_member where memberid =(select max(memberid)from vadhuvar_member)"; $result=mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($result)) { $memberid = $row['memberid']; $fname = $row['fname']; $lname = $row['lname']; $gender = $row['gender']; $add1 = $row['add1']; $add2 = $row['add2']; $city = $row['city']; $state = $row['state']; $zip = $row['zip']; $country = $row['country']; //$photo1=chunk_split(base64_encode($photo1)); $photo1 = $row['photo1']; //echo base64_encode($photo1); echo "Name:$fname $lname<br>"; //echo "Name:$_POST[fname]$_POST[lname]<br>"; echo "Gender:$gender<br>"; echo "Address:$add1 $add2<br>"; echo "$city $state $zip<br>"; echo "$country<br>"; //echo "<img src=".upload/$photo1 ." alt="" width="200" height="300" /><br />"; } //} ?> <!--img src="<?php echo '.upload/$_FILES[image][name]?memberid=$row[memberid]'?>"alt="" width="200" height="300"/-->; <?php echo "<img src=". upload/$row[photo1]" alt="" width="200" height="300">";?> <!--img src="<?php echo ".\upload\$photo1" ?>" alt="" width="200" height="300"/-->; <? set_magic_quotes_runtime(1); // turn back on ?> Thanks Smita You know... if you run that script, it should be spitting a bunch of errors at you. Try turning error_reporting on and that should be the key. Quote Link to comment Share on other sites More sharing options...
smitagodbole Posted August 28, 2010 Author Share Posted August 28, 2010 error reporting is already on. It is showing me such as..... ; Default Value: E_ALL & ~E_NOTICE ; Development Value: E_ALL & ~E_NOTICE?(I hope this is correct) Actually, I am seeing all other info such as name,gender and address etc except image. I am seeing empty frame for photo. Thanks Smita Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted August 28, 2010 Share Posted August 28, 2010 Post the code you're currently using. Quote Link to comment Share on other sites More sharing options...
litebearer Posted August 28, 2010 Share Posted August 28, 2010 also look closely at your use of quotes ie <?php echo "<img src=". upload/$row[photo1]" alt="" width="200" height="300">";?> Quote Link to comment Share on other sites More sharing options...
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