你的 SQL 语法错误？

Question

我有这个函数，其中有一个非常简单的 SQL，它在没有语法的地方给出了错误。

function createPetition($conn, $pName, $pTarget, $pDescription, $userId) {
    $sql = "INSERT INTO petitions (petitionName, petitionTarget, petitionDescription, usersId) VALUES (?, ?, ?, ?);";
    //$sql2 = "INSERT INTO petitions (`petitionDate`) VALUES (NOW());";
    $stmt = mysqli_stmt_init($conn);
    if (!mysqli_stmt_prepare($stmt, $sql)) {
        header("location: ./index.php?error=stmtfailed");
        exit();
    }

    /*if (!mysqli_stmt_prepare($stmt, $sql2)) {
    header("location: ./index.php?error=stmtdatefailed");
    exit();
    }*/
    
    mysqli_stmt_bind_param($stmt, "ssss", $pName, $pTarget, $pDescription, $userId);

    if (!mysqli_query($conn, $sql)) {
        echo "Error description: " . mysqli_error($conn);
    } else {
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);
        header("location: ./petitionfinal.php?error=none");
        exit();
    }
}

其结果是

Error description: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '?, ?, ?, ?)' at line 1

明确无误时

Answer 1

备注

您可能不应该在函数内部使用 exit：return 是更好的选择。
当您使用绑定参数时，您不能在 query 和 prepare 之间 flip-flop。该参数仅在 execute 语句期间绑定。
如果使用对象样式 methods 与过程样式 functions.
相比，数据库交互会容易得多

问题

您的最终 if 语句有 mysqli_query，它不适用于 prepared 语句和 bound 变量。

它尝试运行字面值$sql 的查询。这就是错误响应中有 ??? 的原因。

代码示例

function createPetition($conn, $pName, $pTarget, $pDescription, $userId){
    $sql = "INSERT INTO petitions (petitionName, petitionTarget, petitionDescription, usersId) VALUES (?, ?, ?, ?);";
    //$sql2 = "INSERT INTO petitions (`petitionDate`) VALUES (NOW());";
    $stmt = mysqli_stmt_init($conn);
    if (!mysqli_stmt_prepare($stmt, $sql)) {
    header("location: ./index.php?error=stmtfailed");
    exit();
    }

    /*if (!mysqli_stmt_prepare($stmt, $sql2)) {
    header("location: ./index.php?error=stmtdatefailed");
    exit();
    }*/
    
    mysqli_stmt_bind_param($stmt, "ssss", $pName, $pTarget, $pDescription, $userId);

    if (!mysqli_stmt_execute($stmt)) {
        echo("Error description: " . mysqli_error($conn));
      }else{     
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);
        header("location: ./petitionfinal.php?error=none");
        exit();
      }

}

首选代码

function createPetition($conn, $pName, $pTarget, $pDescription, $userId){
    $sql   = "
        INSERT INTO petitions
        (petitionName, petitionTarget, petitionDescription, usersId)
        VALUES
            (?, ?, ?, ?)
    ";

    $query = $conn->prepare($sql);
    $query->bind_param("ssss", $pName, $pTarget, $pDescription, $userId);

    if($query->execute()){
        return TRUE;
    }
    else{
        // return $mysqli->error;
        return false;
    }
}

// You should carry out error reporting or re-directs outside of your function.

/*

For example:


$makePetition = createPetition($conn, $pName, $pTarget, $pDescription, $userId);

if($makePetition === TRUE){
    header("location: ./petitionfinal.php?error=none");
}
else{
    header("location: ./index.php?error=stmtfailed");
}
*/

Answer 2

问题是你的代码中有一个流氓 mysqli_query()。您需要将其删除。

您可能错过了它，因为您的代码太多了。你应该删除大部分。你真的不需要全部。

代码应如下所示：

function createPetition(mysqli $conn, $pName, $pTarget, $pDescription, $userId) {
    $sql = "INSERT INTO petitions (petitionName, petitionTarget, petitionDescription, usersId) VALUES (?, ?, ?, ?);";
    
    $stmt = $conn->prepare($sql);
    $stmt->bind_param("ssss", $pName, $pTarget, $pDescription, $userId);
    $stmt->execute();

    header("location: ./petitionfinal.php?error=none");
    exit();
}

但是，在函数内部进行重定向并不是一个好主意。我强烈建议把它从功能中取出来。

function createPetition(mysqli $conn, $pName, $pTarget, $pDescription, $userId) {
    $sql = "INSERT INTO petitions (petitionName, petitionTarget, petitionDescription, usersId) VALUES (?, ?, ?, ?);";
    
    $stmt = $conn->prepare($sql);
    $stmt->bind_param("ssss", $pName, $pTarget, $pDescription, $userId);
    $stmt->execute();
}

// when you call
createPetition($conn, $pName, $pTarget, $pDescription, $userId);

header("location: ./petitionfinal.php?error=none");
exit();

看看它有多好。现在更容易发现错误。

如果您想知道错误是怎么回事，那么我必须向您指出 How to get the actual mysql error and fix it?

你的 SQL 语法错误？

You have an error in your SQL syntax error?

php

mysqli