Question

我试图通过一种表单更新2个表，该表单提交到部件表，然后从部件插入记录中获取ID，并在job_parts表中创建一个条目。

我已经尝试过各种论坛的几种选择，但到目前为止还没有运气，我的代码与数据库结构一起在下面。

更新：代码已按照建议进行编辑，但仅将数据发布到“部件”表中，而并没有将数据发布到“ job_parts表”中

表：！（https://drive.google.com/file/d/11I9HZrjc834_Ft5rqZoa0uFoJyDmMWGL/view?usp=sharing）

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

    $job_id = $_POST['job_id'];
    $partName = $_POST['partName'];
    $partCost = $_POST['partCost'];
    $partRetail = $_POST['partRetail'];
    $partQuantity = $_POST['partQuantity'];

    $sql1 = "INSERT INTO parts (part_name, part_cost, part_rrp) VALUES ('$partName', '$partCost', '$partRetail');";
    $sql1 .= "SET @last_id_parts = LAST_INSERT_ID();";
    $sql1 .= "INSERT INTO job_parts (job_id, part_id, quantity) VALUES ('$job_id', @last_id_parts, '$partQuantity')";


    $outcome = mysqli_multi_query($conn, $sql1);
    if ($outcome) {
        do {
            // grab the result of the next query
            if (($outcome = mysqli_store_result($mysqli)) === false && 
 mysqli_error($mysqli) != '') {
                echo "Query failed: " . mysqli_error($mysqli);
            }
        } while (mysqli_more_results($mysqli) && 
 mysqli_next_result($mysqli)); // while there are more results
    } else {
        echo "First query failed..." . mysqli_error($mysqli);
    }

}

Answer 1

此方法应该可以正常工作

    $sql1 = "INSERT INTO parts (part_name, part_cost, part_rrp) VALUES ('$partName', '$partCost', '$partRetail')";
    $result1=mysqli_query($con,$sql1);// where $con is connection string
    $last_id = mysqli_insert_id($con);// where $con is connection string
    $sql2 = "INSERT INTO job_parts (job_id, part_id, quantity) VALUES ('$job_id', $last_id , '$partQuantity')";
  $result2=mysqli_query($con,$sql2);// where $con is connection string

Answer 2

尝试在每个语句后添加;。

$sql1 = "INSERT INTO parts (part_name, part_cost, part_rrp) VALUES ('$partName', '$partCost', '$partRetail');
    SET @last_id_parts = LAST_INSERT_ID();
    INSERT INTO job_parts (job_id, part_id, quantity) VALUES ('$job_id', @last_id_parts, '$partQuantity');";

然后将mysqli_query()更改为mysqli_multi_query()。

Answer 3

这可能就是我写的方式。（“可能是因为我很少使用 mysqli ”）

$conn->begin_transaction();

$stmt1 = $conn->prepare("INSERT INTO parts (part_name, part_cost, part_rrp) VALUES (?, ?, ?");
$stmt1->bind_param("sss", $partName, $partCost, $partRetail);
$stmt1->execute();

$stmt2 = $conn->prepare("INSERT INTO job_parts (job_id, part_id, quantity) VALUES (?, ?, ?)");
$stmt2->bind_param("sis", $job_id, $conn->insert_id, $partQuantity);
$stmt2->execute();

$conn->commit();

请注意，对于数字，您可以将参数与i绑定为整数，将d绑定为双精度（浮点数），而不是s来绑定字符串。例如。如果$job_id是整数，则应输入：

$stmt2->bind_param("iis", $job_id, $conn->insert_id, $partQuantity);

关于错误处理，我建议阅读这篇文章：how-to-get-mysqli-error-information-in-different-environments

还请注意，带参数的预准备语句不仅出于安全原因也是不错的。如果您尝试以自己的方式（... VALUES(.., '$lastName', ..)插入有效名称，例如“ O'Conner”，则查询将失败。

Answer 4

以下内容效果很好

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

    $job_id = $_GET['id'];
    $partName = $_POST['partName'];
    $partCost = $_POST['partCost'];
    $partRetail = $_POST['partRetail'];
    $partQuantity = $_POST['partQuantity'];

     $sql1 = "INSERT INTO parts (part_name, part_cost, part_rrp) VALUES ('$partName', '$partCost', '$partRetail')";
$result1=mysqli_query($conn,$sql1);// where $con is connection string
$last_id = mysqli_insert_id($conn);// where $con is connection string
$sql2 = "INSERT INTO job_parts (job_id, part_id, quantity) VALUES ('$job_id', $last_id , '$partQuantity')";

$ result2 = mysqli_query（$ conn，$ sql2）; //其中$ con是连接字符串

作为以下内容的一部分

<script>
        function openForm() {
          document.getElementById("myForm").style.display = "block";
        }

        function closeForm() {
          document.getElementById("myForm").style.display = "none";
        }
        </script>


        <button class="open-button" onclick="openForm()">Add Part</button>

        <div class="form-popup" id="myForm">
          <form action="" class="form-container" method="POST">
            <h3>Add part</h3>



            <div class="divTable">
                <div class="divTableBody">
                    <div class="divTableRow">
                        <div class="divTableCell"><label for="name"><strong>Name/Description</strong></label></div>
                        <div class="divTableCell"><input id="partName" name="partName" required="" type="text" placeholder="Enter description" /></div>
                    </div>
                    <div class="divTableRow">
                        <div class="divTableCell"><label for="cost"><strong>Cost Price </strong></label></div>
                        <div class="divTableCell"><input id="partCost" name="partCost" required="" type="text" placeholder="Enter cost price" /></div>
                    </div>
                    <div class="divTableRow">
                        <div class="divTableCell"><label for="retail"><strong>Retail Price </strong></label></div>
                        <div class="divTableCell"><input id="partRetail" name="partRetail" required="" type="text" placeholder="Enter retail price" /></div>
                    </div>
                    <div class="divTableRow">
                        <div class="divTableCell"><label for="quantity"><strong>Quantity </strong></label></div>
                        <div class="divTableCell"><input id="partQuantity" name="partQuantity" required="" type="text" placeholder="Enter quantity" /></div>
                    </div>
                    <div class="divTableRow">
                        <div class="divTableCell"><button type="submitpart" name="submitpart" class="btn" onClick="alert('Added!')">Add</button></div>
                        <div class="divTableCell"> <button type="button" class="btn cancel" onclick="closeForm()">Close</button></div>
                    </div>
                </div>
            </div>


          </form>
        </div>

更新一个提交中从第一张表到第二张表的ID的多个表

4 个答案: