使用 PHP 从 MySQL 数据库中获取价值

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时间:2020-08-25 08:31:08  来源:igfitidea点击:

get value from MySQL database with PHP

phpmysqldatabase

提问by Hristo

$from = $_POST['from'];
$to = $_POST['to'];
$message = $_POST['message'];

$query  = "SELECT * FROM Users WHERE `user_name` = '$from' LIMIT 1";
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $fromID = $row['user_id'];
} 

I'm trying to have $formID be the user_id for a user in my database. Each row in the Users table is like:

我试图让 $formID 成为我数据库中用户的 user_id 。用户表中的每一行都类似于:

user_id | user_name | user_type
   1    |  Hristo   |   Agent

So I want $from = 1but the above code isn't working. Any ideas why?

所以我想要,$from = 1但上面的代码不起作用。任何想法为什么?

回答by Sarfraz

Try this:

尝试这个:

$from = mysql_real_escape_string($_POST['from']);
$to = mysql_real_escape_string($_POST['to']);
$message = mysql_real_escape_string($_POST['message']);

$query  = "SELECT * FROM Users WHERE user_name = '$from' LIMIT 1";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_assoc($result)) {
    $fromID = $row['user_id'];
}

Also, make sure that:

另外,请确保:

  • You have connected to the database
  • You do get data from the post, try var_dumpwith your vars eg var_dump($from)
  • 您已连接到数据库
  • 您确实从帖子中获取数据,请尝试var_dump使用您的变量,例如var_dump($from)

回答by CrossProduct

Use mysql_fetch_assoc instead

使用 mysql_fetch_assoc 代替

回答by Babiker

while($row =mysql_fetch_assoc($result)){ $fromID = $row['user_id']; }

while($row =mysql_fetch_assoc($result)){ $fromID = $row['user_id']; }

回答by Your Common Sense

though it should. try this code

虽然应该。试试这个代码

$from    = mysql_real_escape_string($_POST['from']);
$to      = mysql_real_escape_string($_POST['to']);
$message = mysql_real_escape_string($_POST['message']);

$query  = "SELECT * FROM Users WHERE `user_name` = '$from' LIMIT 1";
$result = mysql_query($query) or trigger_error(mysql_error().$query);
$row = mysql_fetch_array($result, MYSQL_ASSOC);
$fromID = $row['user_id'];
echo $fromID;

if it will throw no errors but still print no id, add this line

如果它不会抛出任何错误但仍然没有打印 id,请添加这一行

var_dump($row);

and post here it's output

并在这里发布它的输出

not that you shouldn't use a user name but user id to address particular user.

并不是说您不应该使用用户名而是使用用户 ID 来称呼特定用户。