php printf doesn't work with

Question

I have php prinf and doesn't work with <li type="1">. It works only with chrome browser, what can i do to make it work with all browsers?

My php code is very long and ill provide shorter version:

sample code is like this:

<ul>
<table>
$result = mysql_query("........",$db);

if(mysql_num_rows($result) > 0)
{
    $myrow = mysql_fetch_array($result);
do
{
    printf ("

<tr><td><li type="1">%s</li></td></tr>",$myrow["var"]);
}
while ($myrow = mysql_fetch_array($result));    
}
else
{
    echo "<p align='center'>error...</p>";
}

</table></ul>

here is screenshot

your `printf` is missing a closing double quote, fix the syntax — Édouard Lopez, Sep 08 '13 at 14:51
**Warning:** mysql extension is [deprecated](http://stackoverflow.com/questions/13944956) as of PHP 5.5.0, and will be removed in the future. Instead, the [MySQLi](http://www.php.net/manual/en/book.mysqli.php) or [PDO_MySQL](http://www.php.net/manual/en/ref.pdo-mysql.php) extension should be used. Please don't use `mysql_*` to develop new code. — bansi, Sep 08 '13 at 14:53
missed closing quotes on printf and the quotes around `1` is not quoted. — bansi, Sep 08 '13 at 14:55
your tag order is also messed up `` is not a valid tag inside `
` — bansi, Sep 08 '13 at 14:57
If the sample code is not formatted the same as the real code, we can't see the problem. THIS code should result in zero output because of the quote mark errors. If the real code does output something, then it _doesn't_ look like this, and you're asking us to look at the wrong problem. — , Sep 08 '13 at 14:58
okey i just want to count names with ul li in printf... is there any way? — Lasha, Sep 08 '13 at 15:33

bansi · Accepted Answer · 2013-09-08T16:53:18.683

from the sample code provided you have messed up the tag order. try this. Note i have moved the tag orders. Hope this is what you want to achieve. too unclear from the question.

<table>
<?php
$result = mysql_query("........",$db);

if(mysql_num_rows($result) > 0)
{
    echo '<tr><td><ul>';
    $myrow = mysql_fetch_array($result);
    do
    {
        printf ("<li type=\"1\">%s</li>",$myrow["var"]);
    }
    while ($myrow = mysql_fetch_array($result)); 
    echo '</ul></td></tr>';
}
else
{
    echo "<p align='center'>error...</p>";
}
?>
</table>

tag order should be. you cannot insert <table> as a child of <ul>. <ul> can have only <li> as child.

<table>
   <tr>
       <td>
           <ul>
               <li>...</li>
           </ul>
       </td>
   </tr>
</table>

Edit:
You may not need the <ul>...</ul> tags. you can do it with simple table. Try this

<?php
$result = mysql_query("........",$db);

if(mysql_num_rows($result) > 0)
{
    echo '<table>';
    $rownum=1; //just a counter here, if have other real value then use it
    $myrow = mysql_fetch_array($result);
    do
    {
        printf ("<tr><td>%d</td><td>%s</td></tr>", $rownum, $myrow["var"]);
        $rownum++;
    }
    while ($myrow = mysql_fetch_array($result)); 
    echo '</table>';
}
else
{
    echo "<p align='center'>error...</p>";
}
?>

and if you want to do it with just ordered list you can do it like the following

<?php
$result = mysql_query("........",$db);

if(mysql_num_rows($result) > 0)
{
    echo '<ol>';
    $myrow = mysql_fetch_array($result);
    do
    {
        printf ("<li>%s</li>", $myrow["var"]);
        $rownum++;
    }
    while ($myrow = mysql_fetch_array($result)); 
    echo '</ol>';
}
else
{
    echo "<p align='center'>error...</p>";
}

Note: the type attribute of <li> is deprecated in HTML 4.01. and is not supported in HTML 5.

okey i understand but is there non html method to do list count? i guess i cant achieve thios with html — Lasha, Sep 08 '13 at 15:14
browsers are too much pardoning with improper tag orders and tag closings. but each browsers implement these corrections on their own way. so some invalid codes works on one browser and not other — bansi, Sep 08 '13 at 15:14
can you provide a sample picture of what you want to achieve, may be screen shot of as it is working in chrome? — bansi, Sep 08 '13 at 15:17

php printf doesn't work with

1 Answers1