以下代码返回如下所示的关联数组
Array ( [1] => Array (
[url] => example.com
[title] => Title.example
[snippet] => snippet.example
) )
$blekkoArray = array();
foreach ($js->RESULT as $item)
{
$blekkoArray[$i]['url'] = str_replace ($find, '', ($item->{'Url'}) );
$blekkoArray[$i]['title'] = ($item->{'url_title'});
$blekkoArray[$i]['snippet'] = ($item->{'snippet'});
$i++;
}
print_r ($blekkoArray);
我如何修改数组,使数组元素不是由1,2,3
等标识的,而是由url(例如)标识的
Array ( [example.com] => Array (
[title] => Title.example
[snippet] => snippet.example
) )
您也可以尝试这个(单线解决方案)
$newArray = array( $old[0]['url'] => array_splice($old[0], 1) );
演示
其他解决方案似乎专注于在之后更改阵列
foreach ($js->RESULT as $item)
{
$blekkoArray[str_replace ($find, '', ($item->{'Url'}))] = array(
'title'=> $item->{'url_title'},
'snip pet' => $item->{'snippet'}
);
}
这应该使数组成为您需要的
非常简单,只需使用url而不是$i
foreach ($js->RESULT as $item)
{
$url = str_replace ($find, '', ($item->{'Url'}) )
$blekkoArray[$url]['title'] = ($item->{'url_title'});
$blekkoArray[$url]['snippet'] = ($item->{'snippet'});
$i++;
}
foreach ($js->RESULT as $item)
{
$url = str_replace ($find, '', ($item->{'Url'}) );
$blekkoArray[$url] = array("title"=>($item->{'url_title'}), "snippet"=>($item->{'snipped'}));
}
考虑您的示例如下。其中$js是您想要修改的数组。
$js = array(
1 => array ( 'url' => 'example.com', 'title' => 'Title.example','snippet' => 'snippet.example'),
2 => array ( 'url' => 'example2.com', 'title' => 'Title.example2','snippet' => 'snippet.example2'),
3 => array ( 'url' => 'example3.com', 'title' => 'Title.example3','snippet' => 'snippet.example3'));
$blekkoArray = array();
// The lines below should do the trick
foreach($js as $rows) { // main loop begins here
foreach($rows as $key => $values) { // access what's inside in every $row by looping it again
if ($key != 'url') {
$blekkoArray[$rows['url']][$key] = $values; // Assign them
}
}
}
print_r ($blekkoArray);
$js数组中有多少元素并不重要,因为它每次只会重复这个过程。
foreach ($arr as $key => $value){
$out[$value['url']] = array_slice($value, 1);
}