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我在将表单发布到数据库时遇到问题。第一部分是从一个表中检查 CustomerID 的查询，这部分工作正常。下一部分是将值发布到不同的表中。表格上一切正常，但没有发布到“订单”表，也没有返回任何错误。我错过了什么？

<?php
    //Insert Into database
    $mysqli = @ new mysqli('127.0.0.1','root',null,'storefront');

    if (mysqli_connect_error())
    {
    echo 'Error: Could not connect to database. Please try again later.';
    exit;
    }

    $mysqli->select_db("storefront")
        or die("<p>Unable to select the database.</p>"
        ."<p>Error code ". mysqli_errno($mysqli)
        .": " . mysqli_error($mysqli)) . "</p>";

    $customerid = isset($_POST['customerid']);
    $custid = $_POST['customerid'];

    if($customerid)
    {
        $result = $mysqli->query("SELECT * FROM customer WHERE customerID = '$custid'");
        $row = $result->fetch_row();

        if (!$row)
        {
            echo "Customer Number not found.".mysql_error();
            exit;
        }
    }
    else
    {

        $tableName = "orders";
        $nullstr = "NULL";

        $SQLstring = "INSERT INTO $tableName VALUES
                ('".$nullstr."','".$customerid."', '".$blue."','".$green."','".$red."')";
        $result = $mysqli->query($SQLstring);   

        if ($result)
        {
            echo $mysqli->affected_rows ."  $tableName database.<br />";

            $mysqli->close();
        }
    }
?>
4

2 回答 2

0

这一行：

$SQLstring = "INSERT INTO $tableName VALUES
            ('".$nullstr."','".$customerid."', '".$blue."','".$green."','".$red."')";

应该是这样的

$SQLstring = "INSERT INTO $tableName VALUES
            ('".$nullstr."','".$custid."', '".$blue."','".$green."','".$red."')";

您的插入在 ELSE 中被调用if($customerid)- 看起来您打算将其作为第一个查询的 else：

    if($customerid)
    {
        $result = $mysqli->query("SELECT * FROM customer WHERE customerID = '$custid'");
        $row = $result->fetch_row();

        if (!$row)
        {
            echo "Customer Number not found.".mysql_error();
            exit;
        }      
        else
        {
            $tableName = "orders";
            $nullstr = "NULL";

            $SQLstring = "INSERT INTO $tableName VALUES
                ('".$nullstr."','".$custid."', '".$blue."','".$green."','".$red."')";
            $result = $mysqli->query($SQLstring);   

            if ($result)
            {
                echo $mysqli->affected_rows ."  $tableName database.<br />";

                $mysqli->close();
            }
        }
    }

$blue、$green 和 $red 没有在任何地方定义，但这并不相关。

也许我只是错过了你在这里意图的想法。如果是这样，请告诉我，我将再次审查并更新。

于 2013-04-19T03:14:39.647 回答
0

尝试更换

$customerid至$custid

从

('".$nullstr."','".$customerid."', '".$blue."','".$green."','".$red."')";

至

('".$nullstr."','".$custid."', '".$blue."','".$green."','".$red."')";
于 2013-04-19T03:16:21.293 回答