onmouseover problem
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onmouseover problem
how can I do to eliminate the link that appears on the status bar when the mouse is on one of the links printed wiht this echo?
I have something like this:
echo "<a href='download.php?file=$dfile' target='_blank'>File</a><br>;"
I tried a lot of things, but unsuceccsfuly.
thakns
I have something like this:
echo "<a href='download.php?file=$dfile' target='_blank'>File</a><br>;"
I tried a lot of things, but unsuceccsfuly.
thakns
- twigletmac
- Her Royal Site Adminness
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Try
window.status displays the text specified in the window status-bar.
?>
Code: Select all
echo "<a href='download.php?file=$dfile' target='_blank' window.status='This is a link to $dfile'>File</a><br>;"?>
I pasted HTML, not PHP. Escape the double-quotes " with \" when you echo them.
Now I am off to bed again and cure my hang-over.
Code: Select all
<?php
echo "
<a href='download.php' target='_blank' onMouseOver="status='This is a link to me'; return true;" onMouseOut="status='This was nada'">blalalal</a>
"
?>I cannot test this but I am pretty sure that you could do it this way:
echo <<<LINK
<a href="link.php?var=varData" onMouseover="window.status='Ding Dong'; return true" onMouseout="window.status=' '; return true">Link</a>
LINK;
I find this much easier, but that is just personal oppinion.
Tell me if it works?
echo <<<LINK
<a href="link.php?var=varData" onMouseover="window.status='Ding Dong'; return true" onMouseout="window.status=' '; return true">Link</a>
LINK;
I find this much easier, but that is just personal oppinion.
Tell me if it works?