PHP将记录插入MySQL DB（正在添加多个记录）(PHP Insert Record into MySQL DB (Multiple recorded are being added))

PHP将记录插入MySQL DB（正在添加多个记录）(PHP Insert Record into MySQL DB (Multiple recorded are being added))

我遇到此代码的问题：

<?php $new_value = 'testing'; $con = mysql_connect("localhost","user","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("my_db", $con); mysql_query("INSERT INTO myDB (myField) VALUES ('$new_value')"); mysql_close($con); ?>

2个问题正在发生：

第1个 - 正在插入2个记录而不是1个

第二个 - $ new_value没有改变，当我只想要1时，它创建了具有相同$ new_value值的多个实例。

I am having problems with this code:

<?php $new_value = 'testing'; $con = mysql_connect("localhost","user","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("my_db", $con); mysql_query("INSERT INTO myDB (myField) VALUES ('$new_value')"); mysql_close($con); ?>

2 issues are happening:

1st - 2 records are being inserted instead of 1

2nd - $new_value is not changed and it's creating multiple instances with the same $new_value value when I only want 1.

最满意答案

在阅读你的评论之后，我认为这与你所追求的类似：

INSERT IGNORE INTO tablename (fields) VALUES(values);

使用INSERT IGNORE而不是INSERT。 如果记录没有复制现有记录，MySQL会像往常一样插入它。 如果记录是重复的，则IGNORE关键字告诉MySQL以静默方式丢弃它而不会产生错误。

编辑或正如Vytautas指出的那样; 如果要替换值，请使用REPLACE而不是INSERT

I think this is something similar to what you are after, after reading your comment:

INSERT IGNORE INTO tablename (fields) VALUES(values);

Use INSERT IGNORE rather than INSERT. If a record doesn't duplicate an existing record, MySQL inserts it as usual. If the record is a duplicate, the IGNORE keyword tells MySQL to discard it silently without generating an error.

edit Or as Vytautas points out; if you want to replace the value use REPLACE in stead of INSERT