PHP syntax error in function, don't know why

Bryan Published at Dev

Bryan

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
}

Utkarsh Dixit

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

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