php - php中的sql更新查询错误

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“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
（28个答案）
两年前关闭。
我正在尝试更新已创建的数据库和表中的数据，但给出的错误是：
更新记录时出错：您的SQL语法有错误；请查看MySQL服务器版本对应的手册，在第1行的“where empID=1”附近使用正确的语法
代码如下：

<!DOCTYPE html>
<?php
if( isset( $_REQUEST['update'] ))
{
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "bc140401259_DB";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$employeeID="";
if ( ! empty($_POST['empID'])){
    $employeeID = $_POST['empID'];
}
$Name="";
if ( ! empty($_POST['name'])){
    $Name = $_POST['name'];
}
$Address="";
if ( ! empty($_POST['address'])){
    $Address = $_POST['address'];
}
$Dateofbirth="";
if ( ! empty($_POST['DOB'])){
    $Dateofbirth = $_POST['DOB'];
}
$Salary="";
if ( ! empty($_POST['salary'])){
    $Salary = $_POST['salary'];
}
$timestamp="";
if ( ! empty($_POST['timeIn'])){
    $timestamp = $_POST['timeIn'];
}

$sql = "UPDATE  Employee SET Name = $Name, Address = $Address, Dateofbirth = $Dateofbirth, Salary = $Salary, Timein = $timestampemp where empID = $employeeID";

if (mysqli_query($conn, $sql)) {
    echo "Record updated successfully";
} else {
    echo "Error updating record: " . mysqli_error($conn);
}

mysqli_close($conn);
}

?>

<html>
<body>

<form action="" method="post">
<h2>Add Employee Form</h2>
  Employee ID: <input type="text" name="empID" >
  <br><br>
  Name: <input type="text" name="name" >
  <br><br>

  Address: <input type="text" name="address" >
  <br><br>

  Date of Birth: <input type="text" name="DOB" >
  <br><br>

  Salary: <input type="text" name="salary" >
  <br><br>

  Time In: <input type="text" name="timeIn" >
  <br><br>

<input type="submit" name="update" value="UPDATE" />

</form>
</body>
</html>

最佳答案

这行有误

$sql = "UPDATE  Employee SET Name = $Name, Address = $Address, Dateofbirth = $Dateofbirth, Salary = $Salary, Timein = $timestampemp where empID = $employeeID";

新代码

        <?php
    if( isset( $_REQUEST['update'] ))
    {
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "bc140401259_DB";

    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $dbname);
    // Check connection
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $employeeID="";
    if ( ! empty($_POST['empID'])){
        $employeeID = $_POST['empID'];
    }
    $Name="";
    if ( ! empty($_POST['name'])){
        $Name = $_POST['name'];
    }
    $Address="";
    if ( ! empty($_POST['address'])){
        $Address = $_POST['address'];
    }
    $Dateofbirth="";
    if ( ! empty($_POST['DOB'])){
        $Dateofbirth = $_POST['DOB'];
    }
    $Salary="";
    if ( ! empty($_POST['salary'])){
        $Salary = $_POST['salary'];
    }
    $timestamp="";
    if ( ! empty($_POST['timeIn'])){
        $timestamp = $_POST['timeIn'];
    }
    $sql = 'UPDATE Employee SET Name = "'.$Name.'", Address = "'.$Address.'", Dateofbirth = "'.$Dateofbirth.'", Salary = "'.$Salary.'", Timein = "'.$timestamp.'" where empID = '.$employeeID;
    //echo $sql; die();

    if (mysqli_query($conn, $sql)) {
        echo "Record updated successfully";
    } else {
        echo "Error updating record: " . mysqli_error($conn);
    }
    mysqli_close($conn);
    }

    ?>

在这一行中，数据类型错误与伴随错误

关于php - php中的sql更新查询错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44986980/

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