Question

我有这个嵌套的json数据数组，我正在尝试将特定数据插入MYSQL数据库。但是我得到一个错误，我只是不知道我的代码有什么问题。抱歉还是php / mysql的新手。任何帮助表示赞赏

这是json数组：

[
{
"title": "â˜… (Blackstar)",
"artist": "David Bowie",
"year": "2016",
"genre": "Jazz",
"media": [
{
"totalDiscs": "1",
"position": "1",
"tracks": [
{
"title": "â˜… (Blackstar)",
"number": "1",
"artists": []
},
{
"title": "'Tis A Pity She Was A Whore",
"number": "2",
"artists": []
},
{
"title": "Lazarus",
"number": "3",
"artists": []
},
{
"title": "Sue (Or In A Season Of Crime)",
"number": "4",
"artists": []
},
{
"title": "Girl Loves Me",
"number": "5",
"artists": []
},
{
"title": "Dollar Days",
"number": "6",
"artists": []
},
{
"title": "I Can't Give Everything Away",
"number": "7",
"artists": []
}
]
}
],
"score": 1
}
]

这是我的代码：

$json = json_decode($result, true);

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "4tracks";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} else {
    //echo "connected <br/>";
}   



$sql = "INSERT INTO tracks (artist_name)
VALUES ('".$json[0]['artist']."')";

    if (array_key_exists('genre',$json[0])){
        $sql = "INSERT INTO tracks (track_genre)
            VALUES ('".$json[0]['genre']."')";

    }

    foreach($json[0]['media'] as $key => $values){


        foreach($values['tracks'] as $key1 => $values1) {
            $sql .= "INSERT INTO tracks (track_name)
                VALUES ('".$values1['title']."')";


        }
}



if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();

这是我在wamp上运行.php时的输出：

错误：INSERT INTO曲目（artist_name）VALUES（＆＃39; David Bowie＆＃39;）;插入   INTO跟踪（track_genre）VALUES（＆＃39; Jazz＆＃39;）;

INSERT INTO track（track_name）VALUES（＆＃39;★（Blackstar）＆＃39;）;

INSERT INTO track（track_name）VALUES（＆＃39;＆＃39;可惜她是A   妓女＆＃39）;

INSERT INTO track（track_name）VALUES（＆＃39; Lazarus＆＃39;）;

INSERT INTO track（track_name）VALUES（＆＃39;苏（或者在一个季节）   犯罪）＆＃39）;

INSERT INTO track（track_name）VALUES（＆＃39; Girl Loves Me＆＃39;）;

INSERT INTO track（track_name）VALUES（＆＃39; Dollar Days＆＃39;）;

INSERT INTO track（track_name）VALUES（＆＃39;我不能给予一切   远＆＃39）;

您的SQL语法有错误;检查手册   对应于您的MySQL服务器版本，以便使用正确的语法   靠近＆＃39; INSERT INTO曲目（track_genre）VALUES（＆＃39; Jazz＆＃39;）; INSERT INTO   曲目（track_n＆＃39;在第2行---

Answer 1

最明显的问题是您正在构建一个包含多个查询的字符串。虽然使用mysqli_执行多个查询是不可能的，但不能使用->query()方法完成，并且可以更加简单地独立执行每个查询。

当您在一个查询中同时将多个列同时插入表中时，您在tracks表中每列编写一个查询。

然后你需要使用许多循环遍历你的JSONdata结构，foreach循环最适合这个目的。

您也使用参数化查询，"title": "'Tis A Pity She Was A Whore"等字符串中的引号问题将自动为您处理。

所以我建议将其作为解决方案

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "4tracks";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    echo "Connection failed: " . $conn->connect_error;
    exit;
}

$j = file_get_contents('tst.json');

$json = json_decode($j);
if (json_last_error() != 0) {
    echo json_last_error_msg();
}

// Notice we prepare the query ONCE, but later execute it many times
// with different data in the parameters

$sql = "INSERT INTO tracks (artist_name, track_genre, track_name) VALUES (?,?,?)";
$stmt = $conn->prepare($sql);
// check the prepare worked, if not report errors and exit
if (! $stmt) {
    echo $conn->error;
    exit;
}
// bind the variables names to the ? place holders
// the variables at this point do not have to exists, or have data in them
$stmt->bind_param('sss', $artist, $genre, $title);


foreach($json as $cd) {

    foreach($cd->media as $media) {

        foreach($media->tracks as $track){

            // load the bound variables with the data for this insert execution
            $artist = $cd->artist;
            $genre = $cd->genre;
            $title = $track->title;

            $result = $stmt->execute();
            // check the insert worked, if not report error
            if (!$result) {
                echo $conn->error;
                exit;
            }
        }
    }
}

Answer 2

INSERT INTO tracks (track_name) VALUES (''Tis A Pity She Was A Whore');

＆＃39; Tis - 你需要逃避那个单引号。

$sql = "INSERT INTO tracks (artist_name)
VALUES ('". addslashes ($json[0]['artist']) ."');";

Answer 3

转义将确保MySQL不会解析任何特殊符号（如JSON中的＆＃34;符号）。

要转义查询，请使用$mysqli->real_escape_string($my_json);。

总是逃避您尝试插入数据库的任何内容。甚至更好 - 使用参数化或准备好的陈述（阅读更多here）。

使用php

3 个答案: