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Thread: setting image names with php

  1. #1

    setting image names with php

    Ok im querying the database and doing if statement on the name of a field, if name is this then set variable image to blank.gif. here is my code:

    PHP Code:
    <?php

    error_reporting
    (E_ALL & ~E_NOTICE);


    require_once(
    "connect2.php");

    $query "SELECT id, title, description, url, type, tutorial FROM design_main.urls WHERE (type = 'tutorial') OR (type = 'article') ORDER BY id DESC LIMIT 5"
    $result mysql_query($query) or die (mysql_error()); 
    $num mysql_num_rows($result);

    if(
    $row[5] = 'article'){
    $image 'article.gif';
    }
    elseif(
    $row[5] = 'photoshop'){
    $image 'ps.gif';
    }
    elseif(
    $row[5] = 'php'){
    $image 'php.gif';
    }

    echo
    "<table width=\"100%\" border=\"0\">";
    if(
    $num 0){
    while(
    $row mysql_fetch_array($resultMYSQL_NUM)){

    echo
    "<tr><td><span class=\"style2\" ><img src=images/$image>&nbsp;<b>$row[1]</b><br>$row[2]<br><i>$row[5]</i><br><br></span></td></tr>";
    }
    }
    else
    {
    echo
    '<span class="style1">The calendar is experiencing technical difficulties. We apologize for the inconvenience.</span>'
    }
    echo
    "</table>";

    ?>
    its hanging up on article and all my images are the article.gif. anyone see what im doing wrong?

  2. #2
    check this:
    if($row[5] = 'article')
    you are not comparing value, which is done with double equal sign :
    if($row[5] == 'article')

    what you are doing on all ifs and elses is assignation, not evaluation...

  3. #3
    remeber, that you're COMPARING, not SETTING variables!
    PHP Code:
    if ($row[5] == 'article') { 
       
    $image 'article.gif'
    } else if (
    $row[5] == 'photoshop') {
       
    $image 'ps.gif';
    } else { 
       
    $image 'php.gif'

    lol - twice the good answer

    [edit]neat code[/edit]

  4. #4
    yeah i had == signs earlier and it still wasnt working for some reason, let me try it again.

  5. #5
    ok i got this now and im getting no images at all

    PHP Code:
    <?php

    error_reporting
    (E_ALL & ~E_NOTICE);


    require_once(
    "connect2.php");

    $query "SELECT id, title, description, url, type, tutorial FROM design_main.urls WHERE (type = 'tutorial') OR (type = 'article') ORDER BY id DESC LIMIT 5"
    $result mysql_query($query) or die (mysql_error()); 
    $num mysql_num_rows($result);

    if(
    $row[5] == 'article'){
    $image 'article.gif';
    }
    elseif(
    $row[5] == 'photoshop'){
    $image 'ps.gif';
    }
    elseif(
    $row[5] == 'php'){
    $image 'php.gif';
    }

    echo
    "<table width=\"100%\" border=\"0\">";
    if(
    $num 0){
    while(
    $row mysql_fetch_array($resultMYSQL_NUM)){

    echo
    "<tr><td><span class=\"style2\" ><img src=images/$image>&nbsp;<b>$row[1]</b><br>$row[2]<br><i>$row[5]</i><br></span></td></tr>";
    }
    }
    else
    {
    echo
    '<span class="style1">The calendar is experiencing technical difficulties. We apologize for the inconvenience.</span>'
    }
    echo
    "</table>";

    ?>

  6. #6
    hi, this bunch of code :

    PHP Code:
    if($row[5] == 'article'){
    $image 'article.gif';
    }
    elseif(
    $row[5] == 'photoshop'){
    $image 'ps.gif';
    }
    elseif(
    $row[5] == 'php'){
    $image 'php.gif';

    must be inside the while { } otherwise when you try to evaluate the value of $row the array has not yet been created.

  7. #7
    ah ok that makes sense. Tried it and it works, thanks again pixel

  8. #8
    good to help you! happy coding

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