PHP syntax error in function, don't know why

Question

I'm getting a syntax error and I can't figure out why. "Parse error: syntax error, unexpected '{' on line 6"

function ExtractCustomField($fieldName, $customFields) {
// $customFields might be an object, NULL, or an array.
$parsed = array();
if (is_array($customFields) == false && $customFields != null) {
$parsed = array($customFields);
} else (is_array($customFields)) {
$parsed = $customFields;
}

// loop through the fields and find the one we are looking for
$returnField = null;
foreach($field as $customFields) {
if ($field->Name == $fieldName) {
$returnField = $field;
break;
}
}

return $returnField
}

You’re missing a semicolon in the return statement.

Gumbo
– Gumbo

2015-01-01 08:47:48 +00:00
Commented Jan 1, 2015 at 8:47 — Gumbo
– Gumbo, Commented Jan 1, 2015 at 8:47

Utkarsh Dixit · Accepted Answer · 2015-01-01 08:50:55Z

1

You forget to put semicolon after $returnField, also use elseif instead of else (else don't need any arguments). Use the code below

function ExtractCustomField($fieldName, $customFields) {
// $customFields might be an object, NULL, or an array.
$parsed = array();
if (is_array($customFields) == false && $customFields != null) {
$parsed = array($customFields);
} elseif (is_array($customFields)) {
$parsed = $customFields;
}

// loop through the fields and find the one we are looking for
$returnField = null;
foreach($field as $customFields) {
if ($field->Name == $fieldName) {
$returnField = $field;
break;
}
}

return $returnField;
}

Hope this helps you

answered Jan 1, 2015 at 8:50

Utkarsh Dixit

4,2853 gold badges18 silver badges42 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Bryan Over a year ago

Thanks, dumb mistake on my part.

Ignacio Vazquez-Abrams · Accepted Answer · 2015-01-01 08:48:10Z

0

else doesn't take any expression after it. Either remove the expression or use elseif instead.

answered Jan 1, 2015 at 8:48

Ignacio Vazquez-Abrams

804k160 gold badges1.4k silver badges1.4k bronze badges

Comments

Tushar Gupta · Accepted Answer · 2015-01-01 08:48:26Z

0

You have missed semicolon in return

return $returnField;
}

Also Else condition else (is_array($customFields)) format is wrong....else dont take any condition after it...You can use Elseif instead

answered Jan 1, 2015 at 8:48

Tushar Gupta

16k1 gold badge33 silver badges53 bronze badges

Comments

exussum · Accepted Answer · 2015-01-01 08:49:05Z

0

else (is_array($customFields)) {

Is invalid. Use either else if or remove the condition

answered Jan 1, 2015 at 8:49

exussum

18.6k8 gold badges35 silver badges67 bronze badges

Comments

Sandipon · Accepted Answer · 2015-01-01 08:50:42Z

0

function ExtractCustomField($fieldName, $customFields) {
// $customFields might be an object, NULL, or an array.
$parsed = array();
if (is_array($customFields) == true && $customFields != null) {
$parsed = array($customFields);
} else {
$parsed = $customFields;
}

// loop through the fields and find the one we are looking for
$returnField = null;
foreach($field as $customFields) {
if ($field->Name == $fieldName) {
$returnField = $field;
break;
}
}

return $returnField;
}

answered Jan 1, 2015 at 8:50

Sandipon

9841 gold badge7 silver badges20 bronze badges

Comments

ahmad rabbani · Accepted Answer · 2015-01-01 08:53:33Z

0

It should else if(is_array($customFields)) or elseif(is_array($customFields))

answered Jan 1, 2015 at 8:53

ahmad rabbani

3192 silver badges9 bronze badges

Comments

Rahul DSouza · Accepted Answer · 2015-01-01 10:08:54Z

I have modified your function, i see there is supposed to be and elseif instead of else and in the foreach loop, the syntax in your code was wrong. I have also assumed that you will be using a single level associative array.

<?php
    function ExtractCustomField($fieldName, $customFields) {
            // $customFields might be an object, NULL, or an array.
            $parsed = array();
            if (is_array($customFields) == false && $customFields != null) {
                $parsed = array($customFields);
            } elseif (is_array($customFields)) {
                $parsed = $customFields;
            }

            // loop through the fields and find the one we are looking for
            $returnField = null;
            foreach($customFields as $field) {
                if ($field == $fieldName) {
                    $returnField = $field;
                    break;
                }
            }
            return $returnField;
        }

        echo ExtractCustomField('name', array('name','emial','mobile','password'));
?>

Collectives™ on Stack Overflow

PHP syntax error in function, don't know why

7 Answers 7

1 Comment

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

7 Answers 7

1 Comment

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related