Question

我正在尝试从'$ old'＆gt;一次交换所有属性ID '$ new'然后保存xml：

$reorder = array( 9=>"8", 8=>"5", 7=>"4", 6=>"3", 5=>"0", 4=>"1", 3=>"9", 2=>"7", 1=>"2", 0=>"6" );

    $objDOM = new SimpleXMLElement(some.xml, null, true);
    foreach ($reorder as $old => $new) {
       $picture = $objDOM->xpath('picture[@id="'.$old.'"]');
       $picture[0]["id"] = $new;
    }
    echo $objDOM->asXML();

以下结果（与Array $ reorder不匹配）

3＆gt; 9
9＆gt; 6
8＆gt; 8
7＆gt; 2
6＆gt; 3
5＆gt; 5
4＆gt; 4
0＆gt; 0
1＆gt; 1
7＆gt; 2

它似乎是按顺序切换id，所以刚刚切换的id然后再次切换，如果它们稍后出现在数组中。

那里我做错了什么？如何让它一次性切换所有id？

...谢谢安迪

Answer 1

答案是两个循环

首先在xpath中搜索旧id，将其存储在数组中然后再次循环以使用新id替换存储的结果

$reorder = array(9 => "8", 8 => "5", 7 => "4", 6 => "3", 5 => "0", 4 => "1", 3 => "9", 2 => "7", 1 => "2", 0 => "6");

$objDOM = new SimpleXMLElement(
      '<pictures>
    <picture id="9">id was 9, should be 8 now</picture>
    <picture id="8">id was 8, should be 5 now</picture>
    <picture id="7">id was 7, should be 4 now</picture>
    <picture id="6">id was 6, should be 3 now</picture>
    <picture id="5">id was 5, should be 0 now</picture>
    <picture id="4">id was 4, should be 1 now</picture>
    <picture id="3">id was 3, should be 9 now</picture>
    <picture id="2">id was 2, should be 7 now</picture>
    <picture id="1">id was 1, should be 2 now</picture>
    <picture id="0">id was 0, should be 6 now</picture>
</pictures>');
$oldPicIds = array();

foreach ($reorder as $old => $new) {
   $oldPicIds[$old] = $objDOM->xpath('picture[@id="' . $old . '"]');
}

foreach ($reorder as $old => $new) {
   $oldPicIds[$old][0]['id'] = $new;
}

echo $objDOM->asXML();

输出：

<?xml version="1.0"?>
<pictures>
    <picture id="8">id was 9, should be 8 now</picture>
    <picture id="5">id was 8, should be 5 now</picture>
    <picture id="4">id was 7, should be 4 now</picture>
    <picture id="3">id was 6, should be 3 now</picture>
    <picture id="0">id was 5, should be 0 now</picture>
    <picture id="1">id was 4, should be 1 now</picture>
    <picture id="9">id was 3, should be 9 now</picture>
    <picture id="7">id was 2, should be 7 now</picture>
    <picture id="2">id was 1, should be 2 now</picture>
    <picture id="6">id was 0, should be 6 now</picture>
</pictures>

要保存数组，您可以使用array_pop来获取最后一次出现的图片@ id = xy。哪个应该是想要的那个（阅读缺点评论）

$reorder = array(9 => "8", 8 => "5", 7 => "4", 6 => "3", 5 => "0", 4 => "1", 3 => "9", 2 => "7", 1 => "2", 0 => "6");

$objDOM = new SimpleXMLElement(
      '<pictures>...</pictures>');

foreach ($reorder as $old => $new) {
   $picture = $objDOM->xpath('picture[@id="' . $old . '"]');
   $picture = array_pop($picture);
   $picture['id'] = $new;
}

echo $objDOM->asXML();

Answer 2

从问题中包含的示例来看，我只是迭代所有<picture/>个元素并相应地更改@id。例如：

foreach ($objDOM->picture as $picture)
{
    $id = (string) $picture['id'];
    $picture['id'] = $reorder[$id];
}

这假设$reorder在文档中使用的每个@id都有一个条目。否则，您需要使用isset()来跳过不需要更改的节点。

php循环不按预期工作

2 个答案: