mikearmas
asked on
I don't think it can be done but i want to ask you all.....
They other day you guess helped me with this php load display a lot of images and have a button the edits the file name. Well it works great one thing is i have over 500 image loaded on the page there small 20k files that fine.
But when i use scrolls down fines a image and renames it by clicking on the edit button the page will reload at the top. So the user has to scroll back down and find where they left off.
Is there any way to fixes this so when they click edit it doesn't reset the page at the top but do it from where they left off?
But when i use scrolls down fines a image and renames it by clicking on the edit button the page will reload at the top. So the user has to scroll back down and find where they left off.
Is there any way to fixes this so when they click edit it doesn't reset the page at the top but do it from where they left off?
<p><h1>
All you have to do if you know the person is type their full name, click edit then move on to the next user. Thank you!
<p></h1>
<?php
echo '<table><tr><td>';
$username = $_SESSION['user'];
$url = $username."./";
$handle = opendir ($url);
while (false !== ($file = readdir($handle))) {
if($file != "." && $file != ".." && $file != basename(__FILE__)) {
echo"<form method=\"post\" action=\"data2file.php\">";
echo '<img border=1 src="'.$file.'">';
echo '<br />';
echo "Enter new name: <input type='text' name='new-name' length='30'>
<textarea type=text size='1' height='1'style='display:none' name='file-name' >".$file."</textarea>";
echo "<input type=submit name=\"$file\" value=Edit>" ;
print $file;
echo '<br>';
echo '<hr>';
echo "</form>";
}
}
echo '</td>';
echo '
</table>';
?>
here is the data2file.php
<html>
<body>
<script type="text/javascript"><!--
setTimeout('Redirect()',1000);
function Redirect()
{
location.href = 'http://laptop-vm/IDbag/edit.php';
}
// --></script>
<?php
echo $_POST['new-name'];
echo '<br>';
echo $_POST['file-name'];
rename($_POST['file-name'], $_POST['new-name'].".jpg");
if (copy("C:/server/xampp/htdocs/IDbag/" .$_POST['new-name'].".jpg","C:/server/xampp/htdocs/IDbag/done/".$_POST['new-name'].".jpg")) {
unlink($_POST['new-name'].".jpg");
}
echo '<br>';
echo "Successfully!";
?>
</body>
</html>
ASKER
ok that helped thank you i got it to loop the id # to each image. if i view the sorce code of the page it does it right
<form method=\"post\" action=\"data2file.php#pho to_"1\"><a id=\"photo_"1"\"><img border=1 src="-1033918356.JPG"></a> <br />Enter new name: <input type='text' name='new-name' length='30>
thats from the source code so it"a right but when i click it this is what i get in the address bar. and a error on the page
http://laptop-vm/IDbag/%5C%22data2file.php?new-name=44532234563&file-name=test.jpg&test.jpg=Edit#photo_%22556\%22
<form method=\"post\" action=\"data2file.php#pho
thats from the source code so it"a right but when i click it this is what i get in the address bar. and a error on the page
http://laptop-vm/IDbag/%5C%22data2file.php?new-name=44532234563&file-name=test.jpg&test.jpg=Edit#photo_%22556\%22
<p><h1>
All you have to do if you know the person is type their full name, click edit then move on to the next user. Thank you!
<p></h1>
<?php
$id = 1;
echo '<table><tr><td>';
$username = $_SESSION['user'];
$url = $username."./";
$handle = opendir ($url);
while (false !== ($file = readdir($handle))) {
if($file != "." && $file != ".." && $file != basename(__FILE__)) {
echo'<form method=\"post\" action=\"data2file.php#photo_'.$id.'\">';
echo '<a id=\"photo_"'.$id.'"\"><img border=1 src="'.$file.'"></a>';
$id++;
echo '<br />';
echo "Enter new name: <input type='text' name='new-name' length='30'>
<textarea type=text size='1' height='1'style='display:none' name='file-name' >".$file."</textarea>";
echo "<input type=submit name=\"$file\" value=Edit>" ;
print $file;
echo '<br>';
echo '<hr>';
echo "</form>";
}
}
echo '</td>';
echo '
</table>';
?>
<form method=\"post\" action=\"data2file.php#phoThat doesn't look that good.to_"1\"><a id=\"photo_"1"\"><img border=1 src="-1033918356.JPG"></a> <br />Enter new name: <input type='text' name='new-name' length='30>
if that is source code / a.k.a. HTML then those \ shouldn't be in there at all ;)
I see u type differently than I do. u use ' to echo. In that case u don't need \" but can just use " in there ;)
Also <a id=\"photo_"1"\"> should be <a id="photo_1">
But I don't see that in the php code u showed beneath that.
So what u need to do is remove all the \ from the code I gave u ;)
if you echo with single '
I see u also sometimes echo with " in that case u either need to leave the \ in there, or also echo with just single '
Dit u follow that ? lol
okay, just copy/paste this part, lol:
echo'<form method="post" action="data2file.php#photo_'.$id.'">';
echo '<a id="photo_'.$id.'"><img border="1" src="'.$file.'"></a>';
echo '<br />';
echo 'Enter new name: <input type="text" name="new-name" length="30">
<textarea type="text" size="1" height="1" style="display: none;" name="file-name">'.$file.'</textarea>';
echo '<input type="submit" name="'.$file.'" value="Edit">';
print $file;
echo '<br>';
echo '<hr>';
echo '</form>';
$id++;
ASKER
yep that worked i didn't get any errors but this is the data2file.php that has a redirect in the code to reload the edit.php page and it's doesn't know about the photo_id #. How can i have to remember it ?
<html>
<body>
<script type="text/javascript"><!--
setTimeout('Redirect()',1000);
function Redirect()
{
location.href = 'http://laptop-vm/IDbag/edit.php';
}
// --></script>
<?php
echo $_POST['new-name'];
echo '<br>';
echo $_POST['file-name'];
rename($_POST['file-name'], $_POST['new-name'].".jpg");
if (copy("C:/server/xampp/htdocs/IDbag/" .$_POST['new-name'].".jpg","C:/server/xampp/htdocs/IDbag/done/".$_POST['new-name'].".jpg")) {
unlink($_POST['new-name'].".jpg");
}
echo '<br>';
echo "Successfully!";
?>
</body>
</html>
ASKER CERTIFIED SOLUTION
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ASKER
thank you
Example piece in HTML:
<a id="photo_30">
<form action="pictures.php#photo
...photo...
<input type="hidden" name="name_orig" value="photo_name" />
<input type="text" name="name_new" value="Add new name here" />
<input type="submit" value="change">
</form>
U see I use photo_30 to reference to that photo. Because u can't use the current photo name, in case u're changing it anyway.
So in your php-loop that shows the photo, just add an increasing number in it.
loop {
echo "<a id=\"photo_".$id."\">";
$id++
}