Question

我似乎在问一个简单的问题，但我找不到适合我的答案。我希望你能帮忙！

我正在编写一个php脚本，它从html文件(name, artist, song)中获取一些表单数据，并将其放入带有(name, artist, song, queue, newcomer)列的表中。现在，列newcomer是一个布尔值;如果传入表单数据的名称值是唯一的，我希望此布尔值从默认值false或0更改为true或1。

这是我的尝试：

$newcomer = mysql_query(
                'IF NOT EXISTS (SELECT DISTINCT name FROM tbl_queue) 
                 THEN UPDATE tbl_queue newcomer=1'
            );

但显然，它不起作用......

提前致谢!!

嗨，谢谢你的回答！它仍然不适合我=（这是来自php的代码：

$ newcomer = mysql_query（“UPDATE tbl_queue SET newcomer = 1 WHERE name ='”。$ _POST [name]。“'AND NOT EXISTS（SELECT DISTINCT name FROM tbl_queue WHERE name ='”。$ _POST [name]。 “'）”）;

Answer 1

"UPDATE tbl_queue SET newcomer=1
    WHERE   name = '" . $name_variable . "' AND
            NOT EXISTS (SELECT DISTINCT name
                            FROM tbl_queue
                            WHERE name='" . $name_variable . "')"

这是你想要做的吗？显然用正确的变量替换$ name_variable。

编辑：重新格式化SQL

Answer 2

我很想做你的初始插入： -

INSERT INTO tbl_queue (name, artist, song, queue, newcomer)
SELECT DISTINCT a.name, a.artist, a.song, a.queue, CASE WHEN b.newcomer IS NULL THEN 0 ELSE 1 END as newcomer
FROM (SELECT '$name' AS name, '$artist' AS artist, '$song' AS song, '$queue' AS queue) a
LEFT OUTER JOIN tbl_queue b
ON a.name = b.name

虚拟选择要插入的值，并根据名称对现有表进行左连接，使用CASE语句检查是否找到匹配的行。

编辑 - 下面是一些显示var赋值和转义值的php。我在你的原帖中使用了mysql_函数，但是真的应该使用mysqli_函数来进行未来验证。

<?php

$name = mysql_real_escape_string($_POST['name']);
$artist = mysql_real_escape_string($_POST['artist']);
$song = mysql_real_escape_string($_POST['song']);
$fcfs_queue = mysql_real_escape_string($_POST['fcfs_queue']);

$sql = ("INSERT INTO tbl_queue (name, artist, song, fcfs_queue, newcomer)
SELECT DISTINCT a.name, a.artist, a.song, a.fcfs_queue, CASE WHEN b.newcomer IS NULL  THEN 0 ELSE 1 END as newcomer
FROM (SELECT '$name' AS name, '$artist' AS artist, '$song' AS song, '$fcfs_queue' AS  fcfs_queue) a
LEFT OUTER JOIN tbl_queue b
ON a.name = b.name");


if (!mysql_query($sql))
{
    die('Error: ' . mysql_error());
}
echo "success";


?>

Answer 3

我希望这是对Kickstart的评论，但它不符合评论的限制，所以..我希望我可以在这里回答吗？

谢谢！然而，这也不起作用...... 它返回此错误：警告：mysql_error（）期望参数1是资源（关于这一行：die（'Error：'。mysql_error（$ sql））;在if语句中）

我的代码是：

$sql = ("INSERT INTO tbl_queue (name, artist, song, fcfs_queue, newcomer)
SELECT DISTINCT a.name, a.artist, a.song, a.fcfs_queue, CASE WHEN b.newcomer IS NULL  THEN 0 ELSE 1 END as newcomer
FROM SELECT '$name' AS name, '$artist' AS artist, '$song' AS song, '$fcfs_queue' AS  fcfs_queue) a
LEFT OUTER JOIN tbl_queue b
ON a.name = b.name");


if (!mysql_query($sql))
  {
  die('Error: ' . mysql_error($sql));
  }
echo "success";

Answer 4

我会使用触发器，因为它是您表的“固定”要求：

DROP TRIGGER IF EXISTS `newComerCheck`//
CREATE TRIGGER `newComerCheck` BEFORE INSERT ON `songs`
 FOR EACH ROW BEGIN
    DECLARE c integer;
    SELECT COUNT(name) FROM `songs` WHERE `name`=NEW.name INTO c;

    IF (c >= 1) THEN
      SET NEW.newcomer=0;
    ELSE 
      SET NEW.newcomer=1;
    END IF;
  END
//

见小提琴： http://sqlfiddle.com/#!2/3e73b5/1

mysql：IF'some columns value'是DISTINCT THEN的反转布尔值

4 个答案: