vikamjain
Last Activity: Jan 28 '10, 03:56 PM
Joined: Jan 25 '10
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As I said before I have given the actual url value in $pic1. which is working fine and when I used $pic which has the same value as $ pic1(which is evident from this fact that both are giving same thing when echo command is used) -
here all the connection to database are done in some other file. This file was included in that file so connection won't be a problem. $pic1 is showing the picture while $pic is not showing the picture when used in the command
echo "<img src='$pic or $pic1' />";Leave a comment:
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Here is my code
...Code:<h2> Here is your Quiz</h2> <?php $i=0; echo "<table border=\"1\" align=\"top\">"; echo "<tr><th>Question no</th>"; echo "<th>Question </th>"; echo "<th>Hint</th>"; echo "<th>Answer</th>"; echo "<th>Picture</th></tr>";
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helllo jorgi I used follwing four kind of pathnames in $s
but none of them workedCode:$s="http://localhost/myfolder/imagefolder/image.jpeg" //with quotations $s="c://xampp/htdocs/myfolder/imagefolder/image.jpeg" $s=http://localhost/myfolder/imagefolder/image.jpeg //without quotations $s=c://xampp/htdocs/myfolder/imagefolder/image.jpeg
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@ Zorgi Sorry, In my question i forgot to put <> but in actual i have tried this command
echo "<image src='$S' width='60' height='80' />";
but it did not work for meLeave a comment:
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How to insert image in php
Hii all
It might be a silly doubt but I am puzzled a lot by it.
Basically my application need to display uploaded file from user.
The name of image file is variable. Lets say the path to uploaded file is stored in
Following four are various combination of $s
...Code:$S. Now when i used following commands $s="http://localhost/myfolder/imagefolder/image.jpeg" //with quotations $s="c://xampp/htdocs/myfolder/imagefolder/image.jpeg"
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