所以我有一个这样的XML:
<cars>
<brand name="Audi">
<model>A1</model>
<model>A3</model>
<model>A5</model>
</brand>
<brand name="Ferrari">
<model>F12</model>
<model>FF</model>
</brand>
</cars>
我想的是将其转换为这样的:$汽车[“奥迪”] [0] [“A1”]等,但我不知道如何让内部文本有入标签(例如:F12 )。 我用的方式simplexml的努力!
所以,现在我这样做:
$doc = new SimpleXmlElement($xml, LIBXML_DTDLOAD);
$brands = $doc->xpath('//brand[@model="Audi"]');
$model_1 = $brands[0]->model[0];
当然还有没有反应?
<?php
$xml = '<cars>
<brand name="Audi">
<model>A1</model>
<model>A3</model>
<model>A5</model>
</brand>
<brand name="Ferrari">
<model>F12</model>
<model>FF</model>
</brand>
</cars>';
$doc = simplexml_load_string($xml);
foreach ($doc->children() as $brand) {
foreach ($brand->children() as $model) {
$cars[(string)$brand->attributes()->name][] = (string)$model;
}
}
echo '<pre>';
print_r($cars);
echo '</pre>';
?>
尝试这个 :
//cars/brand[@name="Audi"]/*[1]
你的错误:
- 属性的匹配应该是
@name="Audi"
-
*[1]
是第一个子节点
例
$models = $doc->xpath('//cars/brand[@name="Audi"]/*[1]');
var_dump((string)current($models));
<cars>
<brand name="Audi">
<model>A1</model>
<model>A3</model>
<model>A5</model>
</brand>
</cars>
$cars = simplexml_load_file("cars.xml"); // root tag cars
// echo $cars->brand[0]['name'];
foreach ($cars->brand[0]->model as $model) {
echo $model;
}
我已经变得更加冷静这个例子:
<?php
echo "<head><style>html,body{padding:0;margin:0;background-color:black;text-align:center;}.ul{border-bottom:10px dashed #555555;width:50%;margin-left:25%;margin-right:25%;list-style-type:none;box-shadow:0px 0px 2px gold;}.li{font-size:100px;background-color:silver;color:white;font-family:arial;text-shadow:1px 1px black;}.li:nth-child(even){background-color:yellow;}</style></head><body>";
$cars = simplexml_load_file("cars.xml"); // root tag cars
// echo $cars->brand[0]['name'];
foreach($cars->brand as $brand) {
echo "<div class='ul'>";
foreach($brand->model as $model) {
echo "<div class='li'>";
echo $model;
echo "</div>";
}
echo "</div>";
}
echo "</body>";
确定这样的伎俩是要转换为字符串标签的时候当然迭代和重复它!
(字符串)$模式;
原来如此! 我认为这是空的,因为调试器未归我任何东西,当我检查它。
您的所有问题正确的工作,非常感谢你!