Need help with gallery script I am writing

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  • A

    Need help with gallery script I am writing

    Hi everyone

    I was working on a php script to display images and I am running into
    problems. My problems are when I execute the query to thumb path it won't
    display the image/s.

    Anyhelp is alway's appreciated

    I have a mysql table with the following structure

    image_id Integer Auto_increment
    title varchar
    descript text
    thumb_path varchar
    imaage_path varchar
    category_id Integer

    and my script

    <html>
    <body>


    <?php

    $link = @mysql_connect( "localhost" , "myuser", "mypass");

    if (!$link) {
    print "Could not connect to server";
    exit;
    }

    if (!@mysql_select _db("photos_db" )) {
    print "Could not select database";
    exit;
    }

    $query1 = "SELECT * FROM categories";
    $query2 = "SELECT * FROM images ORDER BY image_id DESC";

    $result1 = mysql_query($qu ery1);
    $result2 = mysql_query($qu ery2);

    ?>

    <table width="700" bgcolor="#CCCCC C">
    <tr>
    <td width="700" align="center" valign="middle" height="50">
    <?php

    if ($result1 && @mysql_num_rows ($result1) > 0) {
    $data1 = "";
    while ($row1 = mysql_fetch_arr ay($result1)) {

    $data1 .= '<font color="black" size="14">';
    $data1 .= $row1['category_descr iption'];

    }
    echo $data1;
    }

    ?>
    </td>
    </tr>


    <?php
    //here is where I am running into problems
    //it does not want to display the image even though I specify the path
    if ($result2 && @mysql_num_rows ($result2) > 0) {
    $data2 = "";
    while ($row2 = mysql_fetch_arr ay($result2)) {
    $data2 .= $row2['thumb_path'];


    ?>


    <td><img src="<?php echo "$data2";?> "></td>

    <?php
    }
    }
    ?>
    </tr>

    </table>

    </body>
    </html>


  • Shawn Wilson

    #2
    Re: Need help with gallery script I am writing

    A wrote:[color=blue]
    >
    > I was working on a php script to display images and I am running into
    > problems. My problems are when I execute the query to thumb path it won't
    > display the image/s.
    >
    > Anyhelp is alway's appreciated
    >
    > I have a mysql table with the following structure
    >
    > image_id Integer Auto_increment
    > title varchar
    > descript text
    > thumb_path varchar
    > imaage_path varchar
    > category_id Integer
    >
    > and my script
    >
    > <html>
    > <body>
    >
    > <?php
    >
    > $link = @mysql_connect( "localhost" , "myuser", "mypass");
    >
    > if (!$link) {
    > print "Could not connect to server";
    > exit;
    > }
    >
    > if (!@mysql_select _db("photos_db" )) {
    > print "Could not select database";
    > exit;
    > }
    >
    > $query1 = "SELECT * FROM categories";
    > $query2 = "SELECT * FROM images ORDER BY image_id DESC";
    >
    > $result1 = mysql_query($qu ery1);
    > $result2 = mysql_query($qu ery2);
    >
    > ?>
    >
    > <table width="700" bgcolor="#CCCCC C">
    > <tr>
    > <td width="700" align="center" valign="middle" height="50">
    > <?php
    >
    > if ($result1 && @mysql_num_rows ($result1) > 0) {
    > $data1 = "";
    > while ($row1 = mysql_fetch_arr ay($result1)) {
    >
    > $data1 .= '<font color="black" size="14">';
    > $data1 .= $row1['category_descr iption'];
    >
    > }
    > echo $data1;
    > }
    >
    > ?>
    > </td>
    > </tr>
    >
    > <?php
    > //here is where I am running into problems
    > //it does not want to display the image even though I specify the path
    > if ($result2 && @mysql_num_rows ($result2) > 0) {
    > $data2 = "";
    > while ($row2 = mysql_fetch_arr ay($result2)) {
    > $data2 .= $row2['thumb_path'];
    >
    > ?>
    >
    > <td><img src="<?php echo "$data2";?> "></td>
    >
    > <?php
    > }
    > }
    > ?>
    > </tr>
    >
    > </table>
    >
    > </body>
    > </html>[/color]

    I could be wrong (I just looked at it quickly) but it looks like you're shoving
    a bunch of thumb_paths for multiple images into a singe <IMG> tag without even
    specifying an image name. Check the resulting html code. I'm guessing you get
    something like:

    <img src="images/thumbs/images/thumbs/images/thumbs">

    If this is the case, include the image name in $data2 and don't use a while loop
    to build data2. Use a while loop to write all your <IMG> tags.

    Shawn
    --
    Shawn Wilson
    shawn@glassgian t.com

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