PHP中的JSON格式

我試圖將 json 轉換為 PHP 中的文本所以我的舊問題已關閉，我沒有得到答案。

這是我的代碼。從 Mysql 獲取：

<?php
    $result_inventory = $con2->query("SELECT inventory FROM players WHERE id = '" .$selectedprofile["id"]."'"); 
                             $inventory_array = [];
        while ($inventory_data = $result_inventory->fetch_assoc()) { 
            $inventory_array[] = $inventory_data;
        }
?>

顯示結果：

                  <?php 
                $obj = json_encode($result_inventory);
                  foreach($obj as $inventory){
                      
                      ?>
                  <tbody>
                    <tr>
                      <td>
                        <div class="d-flex px-2 py-1">
                          <div>
                            <img src=""  class="avatar avatar-sm me-3" alt="xd">
                          </div>
                          <div class="d-flex flex-column justify-content-center">
                            <h6 class="mb-0 text-sm"></h6>
                          </div>
                        </div>
                      </td>
                      <td>
                        <div class="text-xs font-weight-bold">
                        
<?php echo $inventory['name'];?>  
   <h6 class="mb-0 text-sm"><?php echo $inventory['slot']; ?></h6>
                </div>
                      </td>
                      <td class="text-xs font-weight-bold">
                        <span class="text-xs font-weight-bold"><?php echo $inventory['type']?></span>
                      </td>    
                      
                      <td class="text-xs font-weight-bold">
                        <span class="text-xs font-weight-bold"><?php echo $inventory['slot']?></span>
                      </td>
                                            
                      <td class="text-xs font-weight-bold">
                        <span class="text-xs font-weight-bold"><?php echo $inventory['amount']?></span>
                      </td>
                     
                    </tr>
                    <tr>
  
                    </tr>
                  </tbody>
                  <?php }?>
                </table>

訊息結果：

Invalid argument supplied for foreach()

我的資料庫：

[{"name":"phone","type":"item","slot":1,"amount":1,"info":[]},{"name":"driver_license","type":"item","slot":2,"amount":1,"info":{"birthdate":"1999-02-02","lastname":"Salem","type":"Class C Driver License","firstname":"Saleh"}},{"name":"id_card","type":"item","slot":3,"amount":1,"info":{"birthdate":"1999-02-02","gender":0,"citizenid":"TLT73227","nationality":"Saudi","lastname":"Salem","firstname":"Saleh"}},{"name":"lockpick","type":"item","slot":4,"amount":1,"info":[]}]

所以我想要這樣的結果：

<a>Name: Phone</a>
<a>Type: item</a>
<a>Slot: 1</a>
<a>Amount: 1</a>

所以我試圖得到結果，請幫我提取。

uj5u.com熱心網友回復：

我現在明白了。我假設每個組態檔 ID 只有 1 行。所以

$result_inventory = $con2->query("SELECT inventory FROM players WHERE id = '" . $selectedprofile["id"] . "'");
$inventory_array = [];
while ($inventory_data = $result_inventory->fetch_assoc()) {
    $inventory_array[] = $inventory_data;
}
$json = $inventory_array[0]['inventory']; // if exists
$obj = json_decode($json, true);

// then continue with
foreach ($obj as $inventory) {
    //...
}

標籤：phpmysqljson

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