Question

我有这个复杂的.pdf，我想在JSON解析它，以便我可以构建PHP。

我想构建下拉列表

dropdown

现在我正在做这样无法正常工作的事情

<select name="select-box">
  <option value="apple">apple</option>
    <option value="banana">banana</option>
    <option value="kiwi">kiwi</option>
</select>

我感兴趣的是$json = '[ { "class":"autocomplete", "name":"autocomplete-1481957691348-preview", "value":"there is nothing like me", "id":"autocomplete-1481957691348-preview", "type":"autocomplete", "tag":"INPUT" }, { "class":"form-control", "name":"file-1481957721195-preview", "value":"", "id":"file-1481957721195-preview", "type":"file", "tag":"INPUT" }, { "select-box":[ { "selectName":"apple", "optionValue":"apple" }, { "selectName":"banana", "optionValue":"banana" }, { "selectName":"kiwi", "optionValue":"kiwi" } ] }, { "class":"form-control", "name":"select-1481957826542-preview", "value":"jadu", "id":"select-1481957826542-preview", "type":"select", "tag":"SELECT" } ]';以上的object

JSON

但我不想单独{ "select-box":[ { "selectName":"apple", "optionValue":"apple" }, { "selectName":"banana", "optionValue":"banana" }, { "selectName":"kiwi", "optionValue":"kiwi" } ] }这个。相反，我想解析整个parse

JSON

Answer 1

您可以将true作为第二个参数传递给json_decode，以将数据作为数组而不是对象返回。然后你可以做以下事情：

// Decode all of the JSON string and store into variable
$decodedJson = json_decode($json, true);

// Retrieve only the select box related data from all of the decoded JSON data
$selectOptions = $decodedJson[2]['select-box'];

echo '<select name="">';

foreach($selectOptions as $selectOption) {
    echo '<option  value="'.$selectOption['selectName'].'">'.$selectOption['selectName'].'</option>';
}

echo '</select>';

此外，我已经复制了您使用“选择名称”所做的事情。每个选项的值作为选择框选项值以及名称。不确定是否要更改它以使用＆＃39; optionValue＆＃39;价值相反，但我认为我提出它以防万一。

希望这有帮助！

Answer 2

你的代码中有一些错误，如：

echo '<select name=''>';

应该是：

echo '<select name="">';

您可以将对象转换为数组并解析它：

<?php 
 $json = '[
   {
      "class":"autocomplete",
      "name":"autocomplete-1481957691348-preview",
      "value":"there is nothing like me",
      "id":"autocomplete-1481957691348-preview",
      "type":"autocomplete",
      "tag":"INPUT"
   },
   {
      "class":"form-control",
      "name":"file-1481957721195-preview",
      "value":"",
      "id":"file-1481957721195-preview",
      "type":"file",
      "tag":"INPUT"
   },
   {
      "select-box":[
         {
            "selectName":"apple",
            "optionValue":"apple"
         },
         {
            "selectName":"banana",
            "optionValue":"banana"
         },
         {
            "selectName":"kiwi",
            "optionValue":"kiwi"
         }
      ]
   },
   {
      "class":"form-control",
      "name":"select-1481957826542-preview",
      "value":"jadu",
      "id":"select-1481957826542-preview",
      "type":"select",
      "tag":"SELECT"
   }
]';

$arrData    =  json_decode($json, true);
  echo '<select name="">';
    foreach($arrData as $objData){
        $for_select = $objData['select-box'];
        if(is_array($for_select)){
            foreach($for_select as $sel){
                echo '<option value="'.$sel['selectName'].'">'.$sel['selectName'].'</option>';
            }
        }
    }
  echo '</select>';

Answer 3

你有一些带引号的错误，如果它不是空的你需要在你的嵌套json中迭代

$arrData    =  json_decode($json);
echo '<select name="Select-box">';
foreach($arrData as $key=>$objData){
    if(!empty($objData->{'select-box'})) {
          foreach($objData->{'select-box'} as $select){

              echo '<option value="'.$select->optionValue.'">'.$select->selectName.'</option>';
          }
    }
}
echo '</select>';

//print
//<select name="select-box"><option value="apple">apple</option><option value="banana">banana</option><option value="kiwi">kiwi</option></select>

live sample

如果您不知道属性名称，因为它会随时更改

$arrData    =  json_decode($json);
echo '<select name="select-box">';
foreach($arrData as $key=>$objData){
    foreach($objData as $data) {
        if(is_array($data)) {
             foreach($data as $select){

                  echo '<option value="'.$select->optionValue.'">'.$select->selectName.'</option>';
             }
        }
    }
}

live sample

Answer 4

$arrData    =  json_decode($json, TRUE);
echo '<select name="">';
foreach($arrData as $key=>$objData){
  foreach($objData['select-box'] as $key2=>$objData2){
                 echo '<option value='.$objData2['optionValue'].'>'.$objData2['selectName'].'</option>';
  }
}
echo '</select>';

如何在PHP中解析嵌套的JSON

4 个答案: