php - 类型定义字符串中的元素数量与绑定(bind)变量的数量不匹配

Question

我的 php 文件中不断出现此错误.. 警告:mysqli_stmt_bind_param() [function.mysqli-stmt-bind-param]:类型定义字符串中的元素数量与第 17 行...中的绑定(bind)变量数量不匹配 这是我的代码 - 我试图读取此表中的所有内容并将每一列(例如用户名或 obo)存储到一个数组中，可以是一个完整的数组，也可以是每个谢谢的单独数组

<?php
    $con = mysqli_connect("*****", "****", "***", "***");

    $username = $_POST["username"];
    $title = $_POST["title"];
    $description = $_POST["description"];
    $location = $_POST["location"];
    $cost = $_POST["cost"];
    $obo = $_POST["obo"];
    $dimmension = $_POST["dimmension"];
    $phone = $_POST["phone"];
    $email = $_POST["email"];
    $image = $_POST["image"];
    $image2 = $_POST["image2"];

    $statement = mysqli_prepare($con, "SELECT username,title,description,location,cost,obo,dimmension,phone,email,image,image2 FROM Postings");
    mysqli_stmt_bind_param($statement, $username,$title,$description,$location,$cost,$obo,$dimmension,$phone,$email,$image,$image2);
    mysqli_stmt_execute($statement);

    mysqli_stmt_store_result($statement);
    mysqli_stmt_bind_result($statement, $username,$title,$description,$location,$cost,$obo,$dimmension,$phone,$email,$image,$image2);

    $response = array();

    while(mysqli_stmt_fetch($statement)){  
        $response[] = $username;
    }

    $response["success"] = true; 
    #echo json_encode($respond);
    echo json_encode($response);
?>

这是我的 php 管理表的图片

Answer 1

您正在调用 Select 语句，但没有绑定(bind)任何内容。删除此行

mysqli_stmt_bind_param($statement, $username,$title,$description,$location,$cost,$obo,$dimmension,$phone,$email,$image,$image2);

并且代码将正常工作。祝你好运。