SQLite3 表不接受 INSERT INTO 语句.表已创建，数据库也已创建，但不会将任何内容传递到其中 - SQLite3 table not accepting INSERT INTO statements. The table is created, and so is the database, but nothing is passed into it

SQLite3 table not accepting INSERT INTO statements. The table is created, and so is the database, but nothing is passed into it

<?php
try
{
        //open the database
        $db = new PDO('sqlite:music.db');
        $db->exec("DELETE from Music;");
        $db->exec("INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Whatd I Say', 'Ray Charles', '1956');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Smells Like Teen Spirit.', 'Nirvana', '1991');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Hey Jude', 'The Beatles', '1968');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Johnny B. Goode', 'Chuck Berry', '1958');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Good Vibrations', 'The Beach Boys', '1966');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Respect', 'Aretha Franklin', '1967');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Whats Going On', 'Marvin Gaye', '1971');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Imagine', 'John Lennon', '1971');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('(I Cant Get No) Satisfaction', 'Rolling Stones', '1965');" .
                    "INSERT INTO Music(Title, Author, ReleaseDate) VALUES ('Like A Rolling Stone', 'Bob Dylan', '1965');");

        //now output the data to a simple html table...
        //example of specifier  --> WHERE First=''Josh''; <--
        print "<table border=1>";
        print "<tr><td>Title</td><td>Author</td><td>Y.O.P.</td></tr>";
        $result = $db->query('SELECT * FROM Music ');
        foreach($result as $row)
        {
            print "<td>".$row['Title']."</td>";
            print "<td>".$row['Author']."</td>";
            print "<td>".$row['ReleaseDate']."</td></tr>";
            }
        print "</table>";
        $db = NULL;
        }
    catch(PDOException $e)
    {
        print 'Exception : '.$e->getMessage();
        }
    ?>

我不确定为什么没有在表格中插入任何东西。文件"音乐.db"存在于正确的路径中。

作为记录，我只能使用SQlite3，不允许使用SQL。PHP 是允许的，SQLite3 也是允许的。

SQLite3 的某些实现将每个执行命令限制为只有一个查询。例如，您不能像以前那样链接 INSERT，但必须为每个插入调用 $db->exec（）。

你看出这两行之间的区别了吗？

"INSERT INTO Music(Title, Author, Release Date) VALUES 
"INSERT INTO Music(Title, Author, ReleaseDate) VALUES 
                                         ^

给你。从列名中删除该空格，您应该没问题。

这是我的音乐内容.db在运行代码之前。

sqlite> .dump
PRAGMA foreign_keys=OFF;
BEGIN TRANSACTION;
CREATE TABLE music (Title text, Author text, ReleaseDate text);
COMMIT;

您当前在问题中发布的代码现在无需在此处修改即可成功运行。