类型定义字符串中的元素数与绑定变量数不匹配

Question

我一直在我的PHP文件中收到此错误。.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...在第17行上的绑定变量数量不匹配这是我的代码-我正在尝试从该表中读取所有内容，并将每列（例如用户名或obo）存储到一个数组中，每个数组可能是一个完整的数组，也可能是单个的数组

<?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我的管理表的图片

Answer 1

您正在调用Select语句，但未绑定任何内容。 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。 祝好运。