Question

我现在已经做了一段时间了，并且无法弄清楚我做错了什么。

这个脚本看起来很简单，并且正在将文件添加到服务器，但它没有将位置插入数据库。我在哪里使用代码的INSERT部分失败了？

<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title>Untitled Document</title>
</head>

<body>
    <form method="post" action="photo-upload2.php?district_id=<?php echo $_REQUEST['district_id']?>" enctype="multipart/form-data">
        <p>
          Please Enter the Candidate's  Name.
        </p>
        <p>
          Candidate's Name:
        </p>
        <input type="text" name="nameMember"/>
        <p>
            Please Upload a Photo of the Candidate in gif or jpeg format. The file name should be named after the Candidate's name.<br>
            If the same file name is uploaded twice it will be overwritten! Maxium size of File is 35kb. </p>
        <p>
          Photo:
        </p>
        <input type="hidden" name="size" value="350000">
        <input type="file" name="photo"> 
        <br/>
        <br/>
        <input TYPE="submit" name="upload" title="Add data to the Database" value="Add Pic"/>
      </form>
</body>
</html>

<?php


// This is the directory where images will be saved
$target = "../images/candidates/";
$target = $target . basename( $_FILES['photo']['name']);

// This gets all the other information from the form
$name=$_POST['nameMember'];
$district_id=$_REQUEST['district_num'];
$pic=($_FILES['photo']['name']);


// Connects to your Database
include('db.php');

//Writes the information to the database
$sql = "INSERT INTO candidate_images SET
name='$name',
image='$pic,
district_id='$district_id'";

// Writes the photo to the server
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{

// Tells you if its all ok
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
}

 else { // if name is blank
    if ($name=='') { echo "Please enter Candidate's name.";

    // Gives and error if it's not ok

    } else echo "Sorry, there was a problem uploading your file $name.";}
?>

Answer 1

插入查询错误，请将查询更改为以下内容：

$sql = "INSERT INTO candidate_images (name,image,district_id) VALUES ('$name','$pic','$district_id')";

mysqli_query($connection, $sql);

有关详情，请参阅：mysqli_query

这里我举例说明如何插入数据库：

<?php
    // Database connection establishment
    $con=mysqli_connect("localhost","username","password","db_name");

    // Check connection
    if (mysqli_connect_errno($con)) {
    echo "MySQL database connection failed: " . mysqli_connect_error();
    }

    $sql = "INSERT INTO candidate_images (name,image,district_id) VALUES ('$name','$pic','$district_id')";

    // Insert data into database
    mysqli_query($con,$sql);

?>

Answer 2

$sql = "INSERT INTO candidate_images (`name`,`image`, `district_id`) VALUES ('$name','$pic','$district_id')";

在此处使用mysql_* OR mysqli_*或PDO函数对数据库执行查询。

Answer 3

实际上你的插入查询是错误的。你只需要纠正它。它会工作正常。只需用以下代码替换您的代码，不要忘记执行保存在$ sql变量中的查询。这也是必要的。如果我的工作不起作用，请使用您自己的类/模块来执行查询

while(input.hasNext()) {

 if(input.hasNextInt()) {
    x = input.nextInt();
    if(x <= 0 || x > 3) {
     // throw exception or whatever your requirements is.
    }

    switch(x){
        case 1:
        System.out.println("You're in Gear 1");
            break;
        case 2:
        System.out.println("Gear 2");
            break;
        case 3:
        System.out.println("Gear3");
    } 
 }




}

正在上载文件，但INSERT未向数据库添加位置

3 个答案: