proper concatenation for if isset in a php variable string for an HTML input value

Question

I have a variable $RepCom if it isset, I want it to be the value of the input RepCom.

I'm trying to figure out the right way to make this work...

$x_RepCom = '<input name="RepCom" type="text" value="'. if(isset($RepCom)){echo $RepCom; }.'">';

Just dumping the variable is fine, but I get a Undefined error if it is not set.

 $x_RepCom = '<input name="RepCom" type="text" value="'.$RepCom.'">';

So how do I put the if isset for the $RepCom into the value of the input?

I tryed...

$x_RepCom = '<input name="RepCom" type="text" value="<?php if(isset($RepCom)){echo $RepCom; }?>">';

But that just dumps the string <?php if(isset($RepCom)){echo $RepCom; }?> into the value.

What am I missing?

John Conde · Accepted Answer · 2012-05-09 15:48:54Z

3

You had the right idea, you just had the wrong implementation.

$RepComPrint = (isset($RepCom)) ? $RepCom : '';
$x_RepCom = '<input name="RepCom" type="text" value="'.$RepComPrint .'">';

answered May 9, 2012 at 15:48

John Conde

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2 Comments

Monty Over a year ago

you lost me with the ? and : in there. :(

John Conde Over a year ago

It's called a ternary operator.

Tony · Accepted Answer · 2012-05-09 15:50:36Z

2

You would probably want to put the value as an variable:

if(!isset($RepCom))
    $RepCom='';

$x_RepCom = '<input name="RepCom" type="text" value="'.$RepCom.'">';

Now if it is not set, it will be given value of an empty string, which will prevent the undefined error.

answered May 9, 2012 at 15:50

Tony

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2 Comments

Monty Over a year ago

what about if I had 50 different variables here? Would I need to do that for each one?

Tony Over a year ago

If you ave an array of variable names you can loop through them and check if(!isset(${$varList[$i]})). However, you can set error reporting to a less strict level to avoid the variable undefined warning.

amindfv · Accepted Answer · 2012-05-09 15:55:52Z

1

Are you a functional programmer? The code you wrote ($x_RepCom = '<input name="RepCom" type="text" value="<?php if(isset($RepCom)){echo $RepCom; }?>">';) reminds me of how it would be implemented in Scheme or Haskell.

The important thing about PHP is that control structures like if it doesn't just return a value; you can't insert it arbitrarily into code.

The way you have to do it in PHP is something like

$x_RepCom = '<input name="RepCom" type="text" value="';
if(isset($RepCom)) { $x_RepCom .= $RepCom; }
$x_RepCom .= '>';

(Written the long way for clarity)

answered May 9, 2012 at 15:55

amindfv

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1 Comment

Monty Over a year ago

I'm very new to programming & php. That's just my way I guess. Thanks for the tip.

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