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首页 > 解决方案 > 使用 mysqli_insert_id 提取最近自动生成的 ID 时，Mysqli/PHP/HTML 重复插入记录

php - 使用 mysqli_insert_id 提取最近自动生成的 ID 时，Mysqli/PHP/HTML 重复插入记录

问题描述

我目前有这个代码可以插入到我的收据表中，

$query = "insert into receipt(petname,receipttype)
                           values('$petname','$receipttype')";

我在其中获取自动生成的 id $receiptid=mysqli_insert_id($db); 并将其发送到下一页。但是，当插入记录时，还会生成另一个（两个相同的内容但另一个自动生成的 ID 被输入到数据库中） http://prntscr.com/kpjff6 [具有唯一 ID 的重复字段示例]

if (isset($_POST['cmdedit'])){

    $petname = mysqli_real_escape_string($db,$_POST['petname']);
    $receipttype =  mysqli_real_escape_string($db,$_POST['receipttype']);

    $query = "insert into receipt(petname,receipttype)
                       values('$petname','$receipttype')";



        $result = mysqli_query($db,$query);
            if (mysqli_query($db,$query)){
            $receiptid=mysqli_insert_id($db);           
            if ($result) {
                            echo "<font color = 'green' > Receipt sucessfully obtained! Page will auto-redirect to order confirmation page in 5 seconds! </font>";
                     header( "refresh:5; url=addorderhomepage.php?animalid=".$animalid."&receiptid=".$receiptid);

            }else{
                echo "<p>Something went wrong! </p>" . mysqli_error($db);
    }
    }
    }


  ?>
       <div class = "topbar">
      <h2> Order Receipt </h2>
     </div>

  <?php
  $query = "select *
                    from catalogue
                    where animalid= " . $animalid;

            $result = mysqli_query($db,$query);
    if ($result){
            while ($rows = mysqli_fetch_array($result))
            {
                    ?>

      <form method = "post" action = "" >
        <table>
            <tr>
            <th> Animal ID</th>
            <td bgcolor="#FFFFFF"> <input type ="text" name = "txtanimalid" value = "<?php echo $rows[0]; ?>" readonly /> </td>
            </tr>

            <tr>
            <th> Animal Name </th>
            <td> <input type ="text" name = "petname" value = "<?php echo $rows[1]; ?>" readonly />  </td>
            </tr>

            <tr>
            <th> Animal Type </th>
            <td> <input type ="text" name = "petname" value = "<?php echo $rows[3]; ?>" readonly />  </td>
            </tr>

              <tr>
            <th> Animal Species </th>
            <td> <input type ="text" name = "petname" value = "<?php echo $rows[2]; ?>" readonly />  </td>
            </tr>


             <tr>
            <th> Animal Description </th>
            <td> <input type ="text" name = "petname" value = "<?php echo $rows[4]; ?>" readonly />  </td>
            </tr>


             <tr>
            <th> Receipt type </th>
            <td> <input type ="text" name = "receipttype" value = "printable" readonly />  </td>
            </tr>   

            <tr>
            <th> Receipt Date  </th>
            <td> <input type ="date" name = "orderdate" value="<?php echo date('Y-m-j'); ?>" readonly="readonly"  </td>
            </tr>



            <tr>
            <th> <br/><br/> </th>
            <td> </td>
            </tr>

            <tr>
            <th> </th>
            <td> <input type ="submit" value = "Obtain receipt" name = "cmdedit" /> </td>
            </tr>

标签： phphtmlmysql

解决方案

可能会发生重复行，因为您使用相同的数据调用 mysqli_query 两次，第一次将响应存储在 $result var 中，第二次在 if 语句的条件内。

   // ... code
   $result = mysqli_query($db,$query);
- if (mysqli_query($db,$query)) {
+ if ($result) {
      $receiptid = mysqli_insert_id($db);           
      if ($result) {
   // .. rest of the code

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