[英]How can I SELECT field FROM table WHERE id=variable?
I have a variable of the logged in user ( $steamid
) that I need to use to select and echo specific fields from the database.我有一个登录用户 (
$steamid
) 的变量,我需要用它来从数据库中选择和回显特定字段。 I am using the following code, but it is working incorrectly.我正在使用以下代码,但它工作不正常。 All database info is correct, the tables, columns, and variables are not misspelled.
所有数据库信息都是正确的,表、列和变量没有拼写错误。 Not sure what I'm doing wrong.
不知道我做错了什么。
<?php
$con=mysqli_connect("private","private","private","private");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT `bananas` FROM `es_player` WHERE `steamid` = '$steamID'";
if ($result=mysqli_query($con,$sql))
{
// Get field information for all fields
while ($fieldinfo=mysqli_fetch_field($result))
{
printf("bananas: %n",$fieldinfo->bananas);
}
// Free result set
mysqli_free_result($result);
}
mysqli_close($con);
?>
No errors are shown, it simply returns "bananas:" with nothing after it.没有显示错误,它只是返回“bananas:”,后面没有任何内容。 I feel like I didn't to it correctly, does anyone know what I might've done wrong?
我觉得我没有正确对待它,有谁知道我可能做错了什么? Here is a screenshot of my database table so you know what it looks like http://puu.sh/gCY3d/983b738458.png .
这是我的数据库表的屏幕截图,因此您知道它的样子http://puu.sh/gCY3d/983b738458.png 。
Try this:尝试这个:
$query = Mysqli_Query($con, "SELECT * FROM `es_player` WHERE `steamid`='$steamID'") or die(mysql_error());
if( ! mysqli_num_rows($query) )
{
echo 'No results found.';
}
else
{
$bananas_array = mysqli_fetch_assoc($query);
$bananas = $bananas_array['bananas'];
echo 'Number of bananas: '. $bananas;
}
If this doesn't work, there is a problem with STEAM_ID format.如果这不起作用,则 STEAM_ID 格式有问题。 You could try triming the IDs to be JUST a number, and add the STEAM_x:x: to it later.
您可以尝试将 ID 修剪为一个数字,然后再添加 STEAM_x:x: 到它。
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