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[SOLVED] adding hyperlink for one of the <td> element


gojakie

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    echo "<table>";
    $handle = fopen($url, "r");
    while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
    {
        echo "<tr>";
        foreach($data as $d)
            echo "<td>$d</td>";
            echo "<td>View</td>";
        echo "</tr>";
    }
    fclose($handle);
    echo "</table>";

 

The output of the above code is a table which is a grid of 100 rows and 12 columns. The last column is the <td>View</td> on which I need to add a hyperlink after clicking on which it should open a small window-like page without toolbars/menubars that should contain an image. The name of the image will be text in the first column.jpg

 

Example

 

A B C D E F G H I J K View

L M N O P Q R S T U V View

 

If I click on the first View, I should get a page with an image embeded in it A.jpg

 

If I click on the first View, I should get a page with an image embeded in it B.jpg

 

Thanks

    echo "<table>";
    $handle = fopen($url, "r");
    while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
    {
        echo "<tr>";
        foreach($data as $d)
            echo "<td>$d</td>";
        echo "<td onclick=\"window.open('{$data[0]}.jpg','','status=0,toolbar=0,location=0,menubar=0,directories=0,width=350,height=250');\">View</td>";
        echo "</tr>";
    }
    fclose($handle);
    echo "</table>";

 

for more on window.open look here: http://www.javascript-coder.com/window-popup/javascript-window-open.phtml

It works. Thanks

 

But the word "View" doesn't look like hyperlink. It looks like a normal font. Even if I put my mouse over it, it the cursor doesn't change to finger, it looks like a arrow. The user will not understand that this is a link. I had put a <u> element to underline it, but it doesn't look like a link. It works though

 

I want to add one more column to my table at the end next to "View". The content of this column would be a small image 15x15 pixel image. There are three images. If the third column has a negative number, the last column should show n.jpg. p.jpg for positive number and z.jpg for zero.

 

Thanks

    echo "<table>";
    $handle = fopen($url, "r");
    while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
    {
        echo "<tr>";
        foreach($data as $d)
            echo "<td>$d</td>";
        echo "<td><a href="{$data[0]}.jpg" onclick=\"window.open('{$data[0]}.jpg','','status=0,toolbar=0,location=0,menubar=0,directories=0,width=350,height=250');return false;\">View</a></td>";
        echo "</tr>";
    }
    fclose($handle);
    echo "</table>";

It gives me an error arse error: syntax error, unexpected '{', expecting ',' or ';'

 

I tried this thrice. I then changed the string to the previous one that you gave which worked and again tried the second solution but the same error.

 

I am also looking solution for the second question where I want to add one more column to my table at the end next to "View". The content of this column would be a small image 15x15 pixel image. There are three images. If the third column has a negative number, the last column should show n.jpg. p.jpg for positive number and z.jpg for zero.

 

Thanks

    echo "<table>";
    $handle = fopen($url, "r");
    while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
    {
        echo "<tr>";
        foreach($data as $d)
            echo "<td>$d</td>";
        echo "<td><a href=\"{$data[0]}.jpg\" onclick=\"window.open('{$data[0]}.jpg','','status=0,toolbar=0,location=0,menubar=0,directories=0,width=350,height=250'); return false;\">View</a></td>";
        echo "</tr>";
    }
    fclose($handle);
    echo "</table>";

 

Needed to escape the quotes around the href= portion.

yeah...forgot to escape the quotes:

    echo "<table>";
    $handle = fopen($url, "r");
    while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
    {
        echo "<tr>";
        foreach($data as $d)
            echo "<td>$d</td>";
        echo "<td><a href=\"{$data[0]}.jpg\" onclick=\"window.open('{$data[0]}.jpg','','status=0,toolbar=0,location=0,menubar=0,directories=0,width=350,height=250');return false;\">View</a></td>";
        if($data[2] < 0)
            echo "<td><img src=\"n.jpg\" alt=\"Negative\"/></td>";
        elseif($data[2] > 0)
            echo "<td><img src=\"p.jpg\" alt=\"Positive\"/></td>";
        else
            echo "<td><img src=\"z.jpg\" alt=\"Zero\"/></td>";
        echo "</tr>";
    }
    fclose($handle);
    echo "</table>";

yeah regarding escaping quotes, even I thought so as the error was self explanatory. Since I am beginner to php and no knowledge of programming and syntax, I was not able to figure out which quote to change/remove.

 

Issue is resolved. Thank you very much.

 

One request. One of my post is still unanswered. The subject is Beginner in PHP - Need modification in a code and topic id is 230455. Mchl was helping me on that but no responses later. Can you guys help? I have posted 4 problems at the last post and looking answers for that.

 

Thanks again.

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