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[SOLVED] PHP: eval() - Parse error: syntax error, unexpected ')', expecting '('


justravis

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This works:

        $gradecode="echo '<fieldset><legend>Grade: </legend>';";
        $gradecode.="db_selectbox('grade', 'SELECT id, name FROM grade WHERE inactive=0 ORDER by id', 0, 1,0,'Choose Your Grade');";
        $gradecode .= "echo \"</fieldset><br /> \n\";";
eval($gradecode);

 

This tells me theres a Parse error: syntax error, unexpected ')', expecting '(' @ arr_selectbox

        $addcode="echo '<label>Address: <input name=address id=address type=text></label><br />';";
        $addcode.="echo '<label>City: <input name=city id=city type=text></label><br />';";

        $addcode.="arr_selectbox('state', $STATES);";

        $addcode.="echo '<label>Zip: <input name=zip id=zip type=text></label><br />';";
        eval($addcode);

 

Any suggestions?

 

Thanks for your time.

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Try replacing the

        $addcode="echo '<label>Address: <input name=address id=address type=text></label><br />';";
        $addcode.="echo '<label>City: <input name=city id=city type=text></label><br />';";

        $addcode.="arr_selectbox('state', $STATES);";

        $addcode.="echo '<label>Zip: <input name=zip id=zip type=text></label><br />';";
        eval($addcode);

With this;

echo '<label>Address: <input name=address id=address type=text></label><br />
<label>City: <input name=city id=city type=text></label><br />';
arr_selectbox('state', $STATES);
echo '<label>Zip: <input name=zip id=zip type=text></label><br />';

Does it work?

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Hi,

 

I'm not sure about this, but won't the $STATES variable be expanded into the string to be evaluated, rather than being evaluated in the eval call. Try using single quotes for the arr_selectbox line instead of double quotes, see what that does:

 

$addcode.='arr_selectbox("state", $STATES);';

 

Note I also switched the quotes around 'state' as well, although you may be able to escape them.

 

Cheers,

Darren.

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