Question

我想在mysql表中将值增加1但在提交后我发现它已经变为11而不是2，这是我的查询：

UPDATE likes_ SET likes = likes + 1 WHERE post_id = '$pid2'

已修改（评论部分的代码）

<?php

$pid2 = $_POST['like'];
if (isset($_POST['like'])) {
  $q3 = mysql_query("SELECT likes FROM likes_ WHERE post_id = '$pid';");
  $lk = mysql_num_rows($q3);
  if ($lk == 0) {
    mysql_query("INSERT INTO likes_ (id, likes, uid, post_id) VALUES (NULL, 1, '$uid', '$pid2');");
  } else {
    mysql_query("UPDATE likes_ SET likes = 'likes' + 1 WHERE post_id = '$pid2'");
  }
}

Answer 1

从您刚插入的代码中，您不能在字段名称周围使用'。

做到：

mysql_query("UPDATE likes_ SET `likes` = `likes` + 1 WHERE post_id = '$pid2'");

我有一个有效的例子，请查看。

CREATE TABLE likes_ (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
likes INT(11) NOT NULL
);
INSERT INTO likes_ (id, likes) VALUES (NULL, 1);
UPDATE `likes_` SET `likes` = (`likes` + 1);

SELECT * FROM `likes_`

喜欢的结果将是2。

Answer 2

我发现问题，插入查询中的var $ pid必须是$ pid2，非常感谢朋友

Answer 3

如果你输入强制转换为int，那么值可以像int一样计算。如果没有字符串值将像11那样组合。

UPDATE likes_ SET likes = CAST(likes AS int) + 1 WHERE post_id = '$pid2'

我认为这段代码也有用。单引号＆＃39;喜欢＆＃39;可能会更改为字符串值。因此它组合11而不是2

mysql_query("UPDATE likes_ SET likes = likes + 1 WHERE post_id = '$pid2'");

Answer 4

您可以尝试使用以下代码解决问题。

UPDATE likes_ SET `likes` = `likes` + 1 WHERE post_id = '$pid2'"

还要确保likes个字段具有int数据类型。

Answer 5

您可以尝试：

$pid2 = $_POST['like'];
if (isset($_POST['like'])) {


$q3 = mysql_query("SELECT likes FROM likes_ WHERE post_id = '$pid';");

$lk = mysql_num_rows($q3);
$lkp = $lk['likes'] + 1;
if ($lk == 0) {
mysql_query("INSERT INTO likes_ (id, likes, uid, post_id) VALUES (NULL, 1, '$uid', '$pid2');");
} else {
mysql_query("UPDATE likes_ SET likes = ".$lkp." WHERE post_id = '$pid2'");
 }
}

在mysql表中将值增加1

5 个答案: