syntax error, unexpected 'if' (T_IF)

Question

I think I've been staring at this for too long that now I'm failing to identify the problem.

Can someone kindly point out where this is going wrong?

Error:

Parse error: syntax error, unexpected 'if' (T_IF) in /Applications/MAMP/htdocs/mipbi/includes/classes/html.php on line 11

Code:

public static function textbox($name, $maxlength=''){
    return "<input type='text' id='$name' name='$name'  maxlength='$maxlength' value='". if (isset($_POST[$name])) { echo $name; } ."' >";
}

The error is generated at the if statement line.

what's the if doing concatenated to a return followed by an echo? — user557846
– user557846, Commented Jul 29, 2013 at 2:05

score 2 · Accepted Answer · 2013-07-29 02:11:57Z

2

You cannot use if statements in such a way. Use the ternary operator for that purpose or have your variable pre-made before the concatenation.

public static function textbox($name, $maxlength=''){
    return "<input type='text' id='$name' name='$name'  
    maxlength='$maxlength' value='". (isset($_POST[$name]) ? $name : '')."' >";
}

Alternative:

public static function textbox($name, $maxlength='')
{
    if( !isset($_POST[$name] ) {
        $myname = $name;
    } else {
        $myname = '';
    }

return "<input type='text' id='$name' name='$name'  
maxlength='$maxlength' value='". $myname ."' >";

}

edited Jul 29, 2013 at 2:11

answered Jul 29, 2013 at 2:06

anon

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Comments

MisterBla · Accepted Answer · 2013-07-29 02:04:47Z

1

You can't use if statements like that.

To do what you want to achieve you'll need to do this:

public static function textbox($name, $maxlength=''){
    return "<input type='text' id='$name' name='$name'  maxlength='$maxlength' value='".(isset($_POST[$name]) ? $name : '')."' >";
}

answered Jul 29, 2013 at 2:04

MisterBla

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Comments

Tucker · Accepted Answer · 2013-07-29 02:05:37Z

1

try it this way:

public static function textbox($name, $maxlength=''){
    return "<input type='text' id='".$name."' name='".$name."'  maxlength='".$maxlength."' value='".( (isset($_POST[$name]) ? $name:'')."' >";
}

answered Jul 29, 2013 at 2:05

Tucker

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Comments

Class · Accepted Answer · 2013-07-29 02:10:07Z

0

public static function textbox($name, $maxlength=''){
    return "<input type='text' id='$name' name='$name'  maxlength='$maxlength' value='". if (isset($_POST[$name])) { echo $name; } ."' >";
}

should be:

public static function textbox($name, $maxlength=''){
    $ret = "<input type='text' id='$name' name='$name'  maxlength='$maxlength' value='";
    if (isset($_POST[$name])) { $ret .= $name; }
    return $ret . "' >";
}

answered Jul 29, 2013 at 2:10

Class

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Comments

score 0 · Accepted Answer · 2013-07-29 03:36:07Z

0

You can't use the if function inside the retuned value string. I would build the string you want to return first, then return it. In my example the string to be returned is $out. Even if the above worked this is easier to read and maintain.

$out= "<input type='text' id='$name' name='$name'  maxlength='$maxlength' value='";
if (isset($_POST[$name])) {
    $out.= $name; 
} 
$out .='" >";
return $out;

edited Jul 29, 2013 at 3:36

answered Jul 29, 2013 at 2:07

user557846

1 Comment

Ro Yo Mi Over a year ago

Can you expand on why/how your proposed solution solves the question?

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