PHP - Parse error: syntax error, unexpected T_IF

Question

Im getting Parse error from:

$uphalfh = if(isset($price30in)) { echo $price30in->textContent;}
if(isset($price30out)) { echo ", " . $price30out->textContent; }

How can I improve this code to avoid Parse error? I understand that I can't use IF statement in $var

When I Echo Variable's its works like a charm and I get the results that I want. But how can I assign the result of Echo Variable's (on Line:10) to $uphalfh

include_once './wp-config.php';
include_once './wp-load.php';
include_once './wp-includes/wp-db.php';

// A name attribute on a <td>
$price30in = $xpath->query('//td[@class=""]')->item( 1);
$price30out = $xpath->query('//td[@class=""]')->item( 2);

// Echo Variable's
if(isset($price30in)) { echo $price30in->textContent;} if(isset($price30out)) { echo ", " . $price30out->textContent; }

// The wrong CODE i tried:
// $uphalfh = if(isset($price30in)) { echo $price30in->textContent;}
// if(isset($price30out)) { echo ", " . $price30out->textContent; }


// Create post object
  $my_post = array();
  $my_post['post_title'] = 'MyTitle';
  $my_post['post_content'] = 'MyDesciprion';
  $my_post['post_status'] = 'draft';
  $my_post['post_author'] = 1;

// Insert the post into the database
$post_id =  wp_insert_post( $my_post );
update_post_meta($post_id,'30_min',$uphalfh);

If you have $price29in, $price28in, etc... you should read up on variable variables or arrays. — Mike B, Oct 14 '13 at 12:06
you got learn php very neatly. Find a documentation about php here — nurakantech, Oct 14 '13 at 12:35

vbrmnd · Answer 1 · 2013-10-14 12:02:31Z

up vote 1 down vote

remove the assignment.

if(isset($price30in)) { echo $price30in->textContent;}
if(isset($price30out)) { echo ", " . $price30out->textContent; }

answered Oct 14 '13 at 12:02

vbrmnd

714413

This code works well:

if(isset($price30in)) { echo "<br />" . $price30in->textContent;} if(isset($price30out)) { echo ", " . $price30out->textContent; }

How can I add the result of this code to $var ? – Cyborg Oct 14 '13 at 12:12

@Cyborg you're going about it the wrong way. Check out my answer. – Mihai Stancu Oct 14 '13 at 12:24

@Cyborg

$varPrice30in = (isset($price30in))?$price30in->textContent:''; $varPrice30out = (isset($price30out))?", " . $price30out->textContent:'';

– vbrmnd Oct 14 '13 at 12:27

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웃웃웃웃웃 · Answer 2 · 2013-10-14 12:02:40Z

up vote 1 down vote

// $uphalfh = you have to complete the variable initialising here and then continue with the conditional statement or remove the variable initialising and continue with the conditional statement because $var = if(conditon){ }else{ }; is wrong syntax in php
if(isset($price30in)) { 
   echo $price30in->textContent;
}
if(isset($price30out)) {
   echo ", " . $price30out->textContent; 
}

answered Oct 14 '13 at 12:02

웃웃웃웃웃

9,405103769

Can you explain reason for downvotting @downvotter. – 웃웃웃웃웃 Oct 14 '13 at 12:05

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Mx Gherkins · Answer 3 · 2013-10-14 12:03:04Z

up vote 1 down vote

$uphalfh = isset($price30in);

if(isset($price30in)) { 
    echo $price30in->textContent;
}
if(isset($price30out)) {
   echo ", " . $price30out->textContent; 
}

answered Oct 14 '13 at 12:03

Mx Gherkins

8,33461756

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Eternal1 · Answer 4 · 2013-10-14 12:02:59Z

up vote 0 down vote

 echo $uphalf = (isset($price30in)) ? $price30in->textContent : "";

answered Oct 14 '13 at 12:02

Eternal1

3,44911429

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Mihai Stancu · Answer 5 · 2013-10-14 14:04:28Z

up vote 0 down vote

The if clause does not return a value, which means you can't use it in the context of assigning a value to variable because there is no value to be had.

$uphalfh = if(isset($price30in)) {
    echo $price30in->textContent;
} if(isset($price30out)) {
    echo ", " . $price30out->textContent;
}

The above code will not work as you expected it, the bellow code should work as expected -- the value of $uphalf will be set to either $price30in->textContent or $price30out->textContent.

if(isset($price30in)) {
    $uphalfh = $price30in->textContent;
} if(isset($price30out)) {
    $uphalfh = ", " . $price30out->textContent;
}

Only if you also want this result to be outputted to the browser (directly visible to a visitor) you could use echo.

if(isset($price30in)) {
    $uphalfh = $price30in->textContent;
}
if(isset($price30out)) {
    $uphalfh = ", " . $price30out->textContent;
}
echo $uphalfh;

edited Oct 14 '13 at 14:04

answered Oct 14 '13 at 12:03

Mihai Stancu

11.6k11741

Please take a look at my updated question. Any suggestions? Thanks – Cyborg Oct 14 '13 at 13:30

My example should work correctly, $uphalfh should be assigned a value based on which of the if conditions is satisfied. – Mihai Stancu Oct 14 '13 at 14:07

@Cyborg did you by any chance intend to have the value of both nodes? Not just one of them? – Mihai Stancu Oct 14 '13 at 14:09

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Cyborg · Accepted Answer · 2013-10-15 15:53:16Z

up vote 0 down vote accepted

A simple edit did the trick! That solved my problem :-)

I replaced:

$uphalfh = if(isset($price30in)) { echo $price30in->textContent;}
if(isset($price30out)) { echo ", " . $price30out->textContent; }

With this:

$uphalfh = $price30in->textContent. ', ' . $price30out->textContent;

edited Oct 15 '13 at 15:53

answered Oct 14 '13 at 14:56

Cyborg

333416

add a comment |

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