php 使用php将数据插入/更新到MySql数据库

Inserting/updating data into MySql database using php

提问by sublimepremise

I am trying to insert/update the MySql database depending on whether a post already exists on the database (I am checking this with a unique user_id). The following works:

我正在尝试根据数据库中是否已存在帖子来插入/更新 MySql 数据库（我正在使用唯一的 user_id 进行检查）。以下工作：

$select_query = "SELECT * ";
$select_query .= "FROM test ";
$select_query .= "WHERE user_id = '$user_id'";

$check_user_id = mysqli_query($connection, $select_query);

$query  = "INSERT INTO test (";
$query .= "  user_id, name, message";
$query .= ") VALUES (";
$query .= "  '{$user_id}', '{$name}', '{$message}'";
$query .= ")";

$result = mysqli_query($connection, $query);

if ($result) {
    echo "Success!";
} else {
    die("Database query failed. " . mysqli_error($connection));
}

However, when I use the following code with an if/else statement, it does not work anymore, although the console reports "Success!" (meaning $result has a value). Any help would be greatly appreciated. Thanks.

但是，当我将以下代码与 if/else 语句一起使用时，它不再起作用，尽管控制台报告“成功！” （意味着 $result 有一个值）。任何帮助将不胜感激。谢谢。

$select_query = "SELECT * ";
$select_query .= "FROM test ";
$select_query .= "WHERE user_id = '$user_id'";

$check_user_id = mysqli_query($connection, $select_query);

if (!$check_user_id) {
    $query  = "INSERT INTO test (";
    $query .= "  user_id, name, message";
    $query .= ") VALUES (";
    $query .= "  '{$user_id}', '{$name}', '{$message}'";
    $query .= ")";
} else {
    $query  = "UPDATE test SET ";
    $query .= "name = '{$name}', ";
    $query .= "message = '{$message}' ";
    $query .= "WHERE user_id = '{$user_id}'";
}

$result = mysqli_query($connection, $query);

if ($result) {
    echo "Success!";
} else {
    die("Database query failed. " . mysqli_error($connection));
}

回答by user2593560

As i understand your code. you are trying to check if the user_id is existing in your database.. i made a simple code and i think its works for me..

据我了解您的代码。您正在尝试检查您的数据库中是否存在 user_id .. 我做了一个简单的代码，我认为它对我有用..

    $select_query = mysql_query("SELECT * FROM test WHERE user_id = '$user_id'") or die (mysql_error());
$result = mysql_num_rows($select_query);

if(!$result){
    $query = mysql_query("INSERT INTO test (user_id, name, message) VALUES ('$user_id', '$name', '$message')");
        if($query){
            echo "Success!";
        }
        else
        {
            die (mysql_error());
        }
}
else{
    $query2 = mysql_query("UPDATE test SET name='$name', message='$message' WHERE user_id = '$user_id'")
}

回答by lejlot

mysql_queryreturns the operation identifier, not the actual result. This is why $check_user_idis always true, so you are always trying to update (even not existing!) rows.

mysql_query返回操作标识符，而不是实际结果。这就是为什么$check_user_id总是如此，所以你总是试图更新（甚至不存在！）行。

you have to "read" the result ofmysql_queryby for example using

你必须“阅读”的结果，mysql_query例如使用

$check_user_id = mysql_num_rows( mysql_query($connection, $select_query) );

now it returns 0 (false) iff there were no results for q $select_query

现在它返回 0 (false) 如果 q 没有结果 $select_query

回答by woofmeow

This statement is giving you a resource to the result

此语句为您提供了结果的资源

$check_user_id = mysqli_query($connection, $select_query);

$check_user_id = mysqli_query($connection, $select_query);

next you are checking for if(!$check_user_id): this condition evaluates to false because of the negation !. Thus your condition goes to the elsepart and and never enters the if.

接下来您要检查if(!$check_user_id)：由于否定，此条件评估为假!。因此，您的条件进入else零件并且永远不会进入if.

The $resultalways has value because you are calling it towards the end of the script.

将$result始终具有价值，因为你在呼唤它向脚本结束。

回答by juanra

Since you previously know the user_id, and assuming that is a primary key in the table, you could use "ON DUPLICATE KEY UPDATE" clause:

由于您之前知道 user_id，并假设它是表中的主键，您可以使用“ON DUPLICATE KEY UPDATE”子句：

$query = mysql_query("INSERT INTO test (user_id, name, message) 
                     VALUES ('$user_id', '$name', '$message')
                     ON DUPLICATE KEY 
                        UPDATE name='$name', message='$message';
");

Same result with only one query.

只有一个查询的相同结果。

Ref: http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html

参考：http: //dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html

回答by Indrajeet Singh

Use Code for data inserting in mysql.

使用代码在mysql中插入数据。

$query = mysql_query("INSERT INTO test set user_id = '$user_id', name = '$name', message = '$message'");
if($query){
    echo "Success!";
}