php遍历插入数据后，前端ajax回调接收不到返回值

Question

html
 <form class="layui-form" action="" method="post" id="frm1" name="frm1">    
<div class="layui-form-item">           
                  <label for="L_email" class="layui-form-label">
                          <span class="x-red"></span>批量id</label>
                      <div class="layui-input-inline" style="width:440px">
                           <textarea id="ids" name="ids" autocomplete="off" class="layui-input" style="width:535px;" readonly ></textarea>        
                      </div>
     </div>
 <div class="layui-form-item">
                      <label for="L_repass" class="layui-form-label">
                      </label>
                      <button type="button" id="adddata" class="layui-btn" lay-submit lay-filter="add" >提交 </button>
                      <button type="reset" onclick="formReset()" class="layui-btn layui-btn-primary">重置</button>
                  </div>
</form>
<script>
//提交
        form.on('submit(add)', function(data){
            $.ajax({ 
                async : false,
                url:'./tijiao_do1.php',   
                method:'post',   
                data:data.field,    
                dataType:'JSON',             
                success: function(res) {
    var successCount = 0;
    var failedCount = 0;
    for (var i = 0; i < res.length; i++) {
        if (res[i].code == '0') {
            successCount++;
        } else {
            failedCount++;
        }
    }
    if (successCount > 0) {
        layer.msg("成功提交" + successCount + "条数据", { icon: 1, time: 1000 }, function() { location.reload(); });
    }
    if (failedCount > 0) {
        alert("提交失败" + failedCount + "条数据");
    }
    $("#adddata").removeAttr("disabled");
},       
                    error:function (data) {
 
                    }     
            }); 
            return false;
        });
</script>

```php
tijiao_do1.php
<?php
  $gain_name=$_POST['gain_name'];
  $ids = $_POST['ids '];
require './connect_db.php';
  
  $ids =explode("\n",$ids );//分隔换行的数据，如果是逗号隔开的则用$sn=explode(',',$sn);
  //$ids =explode(PHP_EOL,$ids );//分隔换行的数据
  for($i=0;$i<count($ids );$i++)
  if(!empty(trim($ids [$i])))   //判断是否为空
   { 
      $sql = "INSERT INTO sev_order (  sn,gain_name) VALUES (  '$ids [$i]', '$gain_name') ";   
       if ($result = $conn->query($sql)) {
        $success[] = array('code' => '0', 'msg' => '发送成功');
    } else {
        $failed[] = array('code' => '-1', 'msg' => '发送失败');
    }
    echo json_encode(array_merge($success, $failed));
   }
  ?>

Answer 1

在您的 PHP 代码中，您将每个值插入数据库时都打印出 JSON 响应。这将导致您向客户端发送多个 JSON 响应，这可能会干扰您的 Ajax 回调函数。您可以在循环之外定义一个 JSON 数组，并将插入操作的结果添加到该数组中，循环结束后打印出整个数组，如下所示：

<?php
$gain_name = $_POST['gain_name'];
$ids = $_POST['ids'];

require './connect_db.php';

$ids = explode("\n", $ids);
$success = array();
$failed = array();

for ($i = 0; $i < count($ids); $i++) {
    if (!empty(trim($ids[$i]))) {
        $sql = "INSERT INTO sev_order (sn, gain_name) VALUES ('$ids[$i]', '$gain_name')";
        if ($result = $conn->query($sql)) {
            $success[] = array('code' => '0', 'msg' => '发送成功');
        } else {
            $failed[] = array('code' => '-1', 'msg' => '发送失败');
        }
    }
}

echo json_encode(array_merge($success, $failed));
?>

此外，您的 Ajax 回调函数检查每个响应的 code 值，以确定操作是否成功。但是，您在 PHP 代码中始终将 code 值设置为 0，这意味着无论操作是否成功，您都会将其视为成功。如果您要在错误发生时发送有意义的错误消息，可以在 else 分支中设置一个不同的 code 值，如下所示：

<?php
$gain_name = $_POST['gain_name'];
$ids = $_POST['ids'];

require './connect_db.php';

$ids = explode("\n", $ids);
$success = array();
$failed = array();

for ($i = 0; $i < count($ids); $i++) {
    if (!empty(trim($ids[$i]))) {
        $sql = "INSERT INTO sev_order (sn, gain_name) VALUES ('$ids[$i]', '$gain_name')";
        if ($result = $conn->query($sql)) {
            $success[] = array('code' => '0', 'msg' => '发送成功');
        } else {
            $failed[] = array('code' => '-1', 'msg' => '发送失败：' . $conn->error);
        }
    }
}

echo json_encode(array_merge($success, $failed));
?>

在这种情况下，如果出现错误，您将在客户端看到一个包含错误消息的 JSON 响应。