更新json结果mysql依次

Question

我实际上想要以相同的顺序更新mysql的输出。我有一个数据库，其中一些数据已经存在，并且sendondate是常见的。我试图更新，但它只更新最后的输出，而不是所有的序列。请帮我从这个..

<?php 
    $data = '[{"message":"Hello+Test+Message","sender":"test","billcredit":"0.00","messageStatus":"DND","sendondate":"2015-04-23 12:27:00","provider":"aaaa"},{"message":"Hello+Test+Message","sender":"test","billcredit":"0.00","messageStatus":"DELIVERD","sendondate":"2015-04-23 12:27:00","provider":"aaaa"}]'; 


     // $objs = json_decode($data); 
    $con=mysqli_connect("localhost","root","Password*9","sms9"); 
     $objs = json_decode($data,true); 
     foreach ($objs as $obj){ 
      $repor = $obj['messageStatus']; 
      // echo $repor . '<br />'; 
      $sen= $obj['sendondate']; 
      //echo $sen; 

      $repor=array(); 
      while($row = mysql_fetch_array($repor)) 
     { 
      $sql2= "Update detail SET Delivery='$repor' WHERE Datetime='$sen'"; 
    if(mysqli_query($con, $sql2)){ 
      echo "up"; 
      } 
     } 

     } 

    ?> 

Answer 1

只需卸下$repor=array()和while条件（我看不到，什么是它的目的）：

<?php 

$data = '[ 
    {"message":"Hello+Test+Message","sender":"test","billcredit":"0.00","messageStatus":"DND","sendondate":"2015-04-23 12:27:00","provider":"aaaa"}, 
    {"message":"Hello+Test+Message","sender":"test","billcredit":"0.00","messageStatus":"DELIVERD","sendondate":"2015-04-23 12:27:00","provider":"aaaa"} 
]'; 

// $objs = json_decode($data); 
$con = mysqli_connect("localhost","root","Password*9","sms9"); 
$objs = json_decode($data,true); 

foreach ($objs as $obj) { 
    $repor = $obj['messageStatus']; 
    // echo $repor . '<br />'; 
    $sen = $obj['sendondate']; 
    //echo $sen; 

    $sql2 = "Update detail SET Delivery='$repor' WHERE Datetime='$sen'"; 
    if (mysqli_query($con, $sql2)) { 
     echo "up"; 
    } 
} 

更新基于您的评论：

在sql中，如果你想修改特定的行，你必须使用适当的条件匹配它们。通常，根据诸如datetime之类的非唯一列来选择行并不是一个好主意。所以，如果你可以改变你都DB和JSON数据，通常的方法是定义某种ID（通常是整数）：

$data = '[ 
    {"id": 123, "message":"Hello+Test+Message","sender":"test","billcredit":"0.00","messageStatus":"DND","sendondate":"2015-04-23 12:27:00","provider":"aaaa"}, 
    {"id": 666, "message":"Hello+Test+Message","sender":"test","billcredit":"0.00","messageStatus":"DELIVERD","sendondate":"2015-04-23 12:27:00","provider":"aaaa"} 
]'; 
... 
    $id = $obj['id']; 
    ... 
    $sql2 = "Update detail SET Delivery='$repor' WHERE id=$id"; 

或者，如果你不能添加额外的列，你必须添加一个已经定义的用于区分具有相同日期时间的行，例如发件人（发件人值必须不同于记录以区分它们，这在json示例数据中不是这种情况，但这仅仅是使用示例）：

... 
    $sender = $obj['sender']; 
    ... 
    $sql2 = "Update detail SET Delivery='$repor' WHERE Datetime='$sen' AND sender='$sender'"; 

额外的安全提示： 一般来说，你应该prope将数据发送到db之前，防止sql injection。或者，您可以使用prepared statements和binded parameters。