Question

我有这个insert.php文件将数据插入我的数据库，但每次提交数据时$result总是返回false。我不知道问题出在哪里，它总是回应"something error"。

##Insert.php

include_once("connection.php");

if (
    isset($_POST['txtCarModel']) && isset($_POST['txtCarType']) && 
    isset($_POST['txtCapacity']) && isset($_POST['image']) &&
    isset($_POST['txtFuelType']) && isset($_POST['txtPlateNumber']) && 
    isset($_POST['txtOwner'])
) {
    $now = DateTime::createFromFormat('U.u', microtime(true));
    $id = $now->format('YmdHis');

    $upload_folder = "upload";
    $path = "ftp://$upload_folder/$id.jpeg";
    $image = $_POST['image'];
    $fullpath = "carkila.esy.es/$upload_folder/$id.jpeg";

    $Car_Model = $_POST['txtCarModel'];
    $Car_Type = $_POST['txtCarType'];
    $Capacity = $_POST['txtCapacity'];
    $Fuel_Type = $_POST['txtFuelType'];
    $PlateNumber = $_POST['txtPlateNumber'];
    $Image = $_POST['image'];
    $Owner = $_POST['txtOwner'];

    $query = "INSERT INTO tbl_cars (Car_Model, owner, Car_Type,Capacity,fuelType,carPlatenuNumber,Image) VALUES ('$Car_Model', '$Owner', '$Car_Type',$Capacity,'$fuelType','$carPlatenuNumber','$Image')"; 

    $result = mysqli_query($conn, $query);

    if ($result > 0) {
        if (isset($_POST['mobile']) && $_POST['mobile'] == "android") {
            echo "success";
            exit;
    }
    echo "Insert Successfully";   
} else {
    if (isset($_POST['mobile']) && $_POST['mobile'] == "android") {
        echo "failed";
        exit;
    }
    echo "Something Error";   
}

Answer 1

$query = "INSERT INTO tbl_cars (Car_Model, owner, Car_Type,Capacity,fuelType,carPlatenuNumber,Image) VALUES ('$Car_Model', '$Owner', '$Car_Type',$Capacity,'$fuelType','$carPlatenuNumber','$Image')";

$ carPlatenuNumber应为$ plateNumber

Answer 2

使用mysqli_affected_rows ()

mysqli_query($conn, $query);
$result = mysqli_affected_rows ($conn);

它将返回受上一操作影响的行数。

mysqli_affected_rows

为什么$ result总是让我错误？

2 个答案: