这是我的general.xml文件: -
<?xml version="1.0" encoding="UTF-8"?>
<results>
<numFound>10</numFound>
<QTime>4</QTime>
<result>
<distance>1071.59873299109</distance>
<name>Irungattukottai</name>
</result>
<result>
<distance>1892.33578431928</distance>
<name>Valapuram</name>
</result>
</results>
region.xml: -
<childrens>
<child_5068 entity_id="5069" value="Irungattukottai" parent_id="4068"/>
<child_5068 entity_id="7140" value="Valapuram" parent_id="4068"/>
</childrens>
product.xml: -
<products>
<product_id value="1">
<tab_id value="351">
<tab_name value="test1"/>
<dist_region value="5069"/>
<dist_region value="5069"/>
<dist_region value="5069"/>
</tab_id>
</product_id>
<product_id value="2">
<tab_id value="352">
<tab_name value="test2"/>
<dist_region value="4457"/>
<dist_region value="7140"/>
<dist_region value="5069"/>
</tab_id>
</product_id>
</products>
我有这三个xml文件我想尝试这样的事情。
在general.xml文件元素<name>Irungattukottai</name>
中
如果此名称在region.xml
退出,则转到entity_id
如果entity_id在product.xml
中退出,则返回product_id
元素属性值...
我试试这段代码: -
<?php
$g = file_get_contents('general.xml');
$r = file_get_contents('region.xml');
$p = file_get_contents('product.xml');
$general = simplexml_load_string($g);
$region = simplexml_load_string($r);
$product = simplexml_load_string($p);
$name = (string)$general->result->name;
if (strlen(trim($name))==0) exit('name not found');
list($entity) = $region->xpath("//*[@value='$name']/@entity_id");
$entity=(string)$entity;
if (strlen(trim($entity))==0) exit('entity_id not found');
list($prid) = $product->xpath("//dist_region[@value='$entity']/ancestor::product_id/@value");
$prid=(string)$prid;
echo "City Name:- $name, Entity_id:- $entity, Product_id:- $prid";
?>
此代码是工作完成但只有一个值,例如: - general.xml
在第一个<name>Irungattukottai</name>
和第二个根<name>Valapuram</name>
中有两个根
但我的代码是工作只获取第一个元素而不是尝试获取wall xml文件...
这是我的正确输出: -
city Name:- Irungattukottai, Entity_id:- 5079, Product_id:- 1
但我想要这种类型的输出: -
city Name:- Irungattukottai, Entity_id:- 5079, Product_id:- 1
city Name:- Valapuram , Entity_id:- 7140, Product_id:- 2
答案 0 :(得分:0)
你需要迭代$ general的数组。
尝试:
echo '<pre>';
print_r($general);
echo '</pre>';
检查是否有一个包含xml文件中所有元素的数组。
答案 1 :(得分:0)
获取general.xml Feed的代码没问题。
您可以将名称命名为
$name1 = (string)$general->result[0]->name;
$name2 = (string)$general->result[1]->name;
or you can use loop to get name one by one.
<?php
foreach($general->result as $row){
if(isset($row->name)){
// do your logic here for region.xml and then for product.xml
}
}
?>
答案 2 :(得分:0)
试试这个:
$xml = simplexml_load_file( 'general.xml' );
$cities = $xml->xpath( '//result/name' );
$cityCount = count( $cities );
if( $cityCount ) {
$region = simplexml_load_file( 'region.xml' );
foreach( $cities as $city ) {
$regions = $region->xpath( '//*[@value="'. ( string ) $city . '"]/@entity_id' );
if( count ( $regions ) ) {
$product = simplexml_load_file( 'products.xml' );
$products = $product->xpath( '//dist_region[@value="' . ( string ) $regions[ 0 ][ 'entity_id' ] . '"]/ancestor::product_id/@value' );
if( count( $products ) ) {
foreach( $products as $p ) {
echo 'city Name:- ' . ( string ) $city . ', Entity_id:- ' . ( string ) $regions[ 0 ][ 'entity_id' ] . ', Product_id:- ' . $p[ 'value' ] . "\n";
}
}
}
}
}
<强>输出强>
city Name:- Irungattukottai, Entity_id:- 5069, Product_id:- 1
city Name:- Irungattukottai, Entity_id:- 5069, Product_id:- 2
city Name:- Valapuram, Entity_id:- 7140, Product_id:- 2
答案 3 :(得分:0)
使用上述两者都可以尝试这样做: -
<?php
foreach($general->result as $row){
if(isset($row->name)){
//create one variable and store in value.
$name = $row->name;
//all you code for region.xml and then for product.xml
}
echo "name:- $name,Entity:- $entity, product:- $prid"."<br>";
}
?>