onclick not working

[ameliabob]

I am building a table at a server and returning it to the user

I have tried putting the $p where it is as shown.  I have also tried it in the first <td

$s =" <td ".$p.">><input type='radio' name='entry' ></td>";

It does not seem to  get to GetFinishData().  Any thoughts or how can I get whatever failure to show up somewhere?

	 $p = "onclick=\"GetFinishData('".$r[rowId]."')\"";
	$s =" <td>><input type='radio' name='entry' .$p.></td>";
	

Thanks

[dsonesuk]

You have not escaped html string with double quotes to insert $p PHP variable.

[ameliabob]

Would you mind modifying whichever string to fix this?

Thanks

 

[dsonesuk]

Everything between double quotes will be treated as text string

" <td>><input type='radio' name='entry' .$p.></td>";

You need to come out text string by exiting out from text string highlighted in red by enclosing within double quotes.

" <td>><input type='radio' name='entry' .$p.></td>";

Now use your little grey cells to figure out where to place these end and start quotes to get "html text string" php code "html text string".

[ameliabob]

OK I tried to simplify the code but still can't get it to work.

	                $s .= "<td";
                $s .= " onclick=\"GetFinishedData('9')\" > ";
                $s .= $r[symbol]."</td>";
	

When I look at the generated code it looks like it should.

Maybe I have run out of little grey cells.

Edited June 4, 2018 by ameliabob
The first line of code should be only $s = "<td";

[iwato]

The term $r[symbol] looks like an unworkable  mixture of PHP and Javascript. 

If $r is the name of a PHP array, then symbol must be enclosed in either single or double quotation marks.  Also,  what is present to cause the appearance of $r[symbol] to be read as the value of an element of a PHP array?

Roddy

[ameliabob]

The whole function looks like this

[code}

        $qry= "select * from holdings where 'pl' is not null order by 'symbol'ASC";
        $result = mysqli_query($cxn,$qry) or die("ShowExit didn't work. ".mysqli_error($cxn));
        if(mysqli_num_rows($result)==0){
            $s = "alertThere are no entries in the database that meet your criteria.  Try changing the search.";
        }else{
            $s = "resp <table id='t01' border='1' >";
            $s .= "<tr><th>Type</th><th>Symbol</th><th>Expire Month</th><th>Qty</th>";
            $s .= "</tr>";
            while($r=mysqli_fetch_array($result)){
                $s .= "<td>".$r[nature]."</td>";
                $s .= "<td";
                $s .= " onclick=\"GetFinishedData('9')\" > ";
                $s .= $r[symbol]."</td>";
                if($r[nature]=="O")
                    $s .= "<td><center>".$r[expireMonth]."</td>";
                else
                    $s .= "<td><center>n/a</td>";
                $s .= "<td>".$r[qty]."</td>";        
                
                $s .= "</tr>";
            }    
            $s .= "</table>";

            echo($s);

[/code]

 

 

Edited June 4, 2018 by ameliabob

[iwato]

When referring to the name of an element in an associative array you must learn to enclose the name of the element in either single or double quotation marks.

Incorrect:  $r[nature]

Correct:  $r['nature']

Roddy