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conan318

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Echo "<img src='http://datenight.netne.net/images/".$info['img'] ."'title='.info['username']'"."' width='50' height='50'>";

 

im trying display the user's, username in the title when the mouse hovers over the img it displays the username of that image.

 

no errors just displays .info as title

 

thanks [/code]

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Echo "<img src='http://datenight.netne.net/images/".$info['img'] ."'title='.$info['username']'"."' width='50' height='50'>";

just noticed i left the $ out but now getting

 

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/a7817347/public_html/login_success.php on line 67

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this is a css related issue right?

try to make your div like this:

float:left;

height:high enough for the multiple images px;

width:max width of the biggest image px; //not sure about this one, what if 2 pics are big enough to fit in the biggest one. That would mean there will be 2 pics horizontal and the rest vertical.

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all the picture are scaled down to 50px by 50px so there all the same.

 

.random_user{

width: 1000px;

height: 50px;

        float:left;

}

 

but still displaying vertical

set the width to 50 and the height to 1000, see if that helps.

 

EDIT: uh wait isn't this what you wanted :S?

is there a way to force the images display vertical across the page instead of horizontal down the page.

 

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so, theres a div tag for each image?

then make a div that contains the newly created divs,

give the main div a float:left;

and make sure all the newly created divs are width:50px; and height:50px;

then let the main div have a longer height and a 50px width.

 

I guess your using a loop of some kind to echo the picture divs?

create an echo of the main div outisde the loop and make it end after the loop has finished then.

this should put all the picture divs inside the main div and would be showed vertical.

 

can I maybe see the loop your using to create the images? I don't quite get what your doing :S

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