Question

请查看：http://codepad.org/uNlMsvwj为json。

$json = json_decode($raw_json);
//print_r($json);
$count = count($json->response->results);

$i = 0;
foreach($json->response->results as $item){

  echo($item->entries[$i]->displayname);echo "<br>";
  echo($item->entries[$i]->location->street);echo " "; echo($item->entries[$i]->location->streetnumber);echo "<br>";
  echo($item->entries[$i]->location->zipcode);echo " ";
  echo($item->entries[$i]->location->city);echo "<br>";echo "<br>";
  echo($item->entries[$i]->phonenumbers[0]->area);echo "/"; echo($item->entries[$i]->phonenumbers[0]->number);echo "<br>";
$i++;
}

问题是只打印了第一个elemet。如果我用1手动更改$ i，我得到第二个。

我现在已经找了3个小时，但找不到解决方案。如果这是初学者的错误，请原谅我。

由于

更新谢谢大家的快速帮助

Answer 1

你的意思是像这样迭代结果吗？

foreach($json->response->results as $item) {
  for ($i = 0; $i < count($item->entries); $i++) {
    echo($item->entries[$i]->displayname);echo "<br>";
    echo($item->entries[$i]->location->street);echo " ";
    echo($item->entries[$i]->location->streetnumber);echo "<br>";
    echo($item->entries[$i]->location->zipcode);echo " ";
    echo($item->entries[$i]->location->city);echo "<br>";echo "<br>";
    echo($item->entries[$i]->phonenumbers[0]->area);echo "/"; 
    echo($item->entries[$i]->phonenumbers[0]->number);echo "<br>";
  }
}

此外，您可以简化echo声明：

foreach($json->response->results as $item) {
  for ($i = 0; $i < count($item->entries); $i++) {
    echo $item->entries[$i]->displayname.'<br>';
    echo $item->entries[$i]->location->street.' ';
    echo $item->entries[$i]->location->streetnumber.'<br>';
    echo $item->entries[$i]->location->zipcode.' ';
    echo $item->entries[$i]->location->city.'<br><br>';
    echo $item->entries[$i]->phonenumbers[0]->area.'/';
    echo $item->entries[$i]->phonenumbers[0]->number.'<br>';
  }
}

Answer 2

你应该像这样回应：

echo $item->entries[$i]->displayname . "<br>";

.连接字符串。此外，echo不使用任何()。

Answer 3

更改

foreach($json->response->results as $item){

到

foreach($json->response->results->entries as $item){

和这些行

echo($item->entries[$i]->displayname);echo "<br>";

到

echo $item->displayname . "<br>";

所以你的代码看起来像是：

foreach($json->response->results->entries as $item){

  echo "{$item->displayname}<br />";
  echo "{$item->location->street} {$item->location->streetnumber}<br />";
  echo "{$item->zipcode}<br />";
  echo "{$item->location->city}<br /><br />";
  echo "{$item->phonenumbers[0]->area}/{$item->phonenumbers[0]->number}<br />";
}

Answer 4

尝试for-loop而不是foreach：

$json = json_decode($raw_json);
//print_r($json);
$count = count($json->response->results);

foreach($json->response->results->entries as $item){
  echo($item->displayname);echo "<br>";
  echo($item->location->street);echo " ";
  echo($item->location->streetnumber);echo "<br>";
  echo($item->location->zipcode);echo " ";
  echo($item->location->city);echo "<br>";echo "<br>";
  echo($item->phonenumbers[0]->area);echo "/"; echo($item->phonenumbers[0]->number);echo "<br>";
}

...或在foreach中包含一个for循环：

$json = json_decode($raw_json);
//print_r($json);
$count = count($json->response->results);

foreach($json->response->results as $item){

  for ($i = 0; $i < count($item->entries); $i++) {

    echo($item->entries[$i]->displayname);echo "<br>";
    echo($item->entries[$i]->location->street);echo " ";
    echo($item->entries[$i]->location->streetnumber);echo "<br>";
    echo($item->entries[$i]->location->zipcode);echo " ";
    echo($item->entries[$i]->location->city);echo "<br>";echo "<br>";
    echo($item->entries[$i]->phonenumbers[0]->area);echo "/";
    echo($item->entries[$i]->phonenumbers[0]->number);echo "<br>";
  }
}

Answer 5

尝试这个，它可行我测试它：

foreach($json->response->results[0]->entries as $item){
  echo($item->displayname);echo "<br>";
  echo($item->location->street);echo " ";
  echo($item->location->streetnumber);echo "<br>";
  echo($item->location->zipcode);echo " ";
  echo($item->location->city);echo "<br>";echo "<br>";
  echo($item->phonenumbers[0]->area);echo "/"; echo($item->phonenumbers[0]->number);echo "<br>";
}

php中的foreach只通过第一个JSON迭代

5 个答案: