Php MySQL查找第一列'；s字符串，回显第二列'；s
首先，抱歉英语不好：(Php MySQL查找第一列'；s字符串，回显第二列'；s,php,mysql,multiple-columns,Php,Mysql,Multiple Columns,首先，抱歉英语不好：( 我想要像steam、origin或uplay这样的密钥系统。我给用户密钥，用户将该代码写入输入，代码将回显用户的代码。 我有1个表。其中有2列。2行对应2列。如下所示： .----------------------. | sifre | bizimkey | |______________________| | A5Sr2A | First Code | |______________________| | FaQ1fS | Scnd. Code | |_
我想要像steam、origin或uplay这样的密钥系统。我给用户密钥，用户将该代码写入输入，代码将回显用户的代码。
我有1个表。其中有2列。2行对应2列。如下所示：
.----------------------.
| sifre | bizimkey |
|______________________|
| A5Sr2A | First Code |
|______________________|
| FaQ1fS | Scnd. Code |
|______________________|
<?php
if ($_POST['pass'] == "A5Sr2A") {
{
echo "First Code";
}
} else {
header('Location:index.html');
}
?>
我尝试了很多代码。我可以这样运行代码：
.----------------------.
| sifre | bizimkey |
|______________________|
| A5Sr2A | First Code |
|______________________|
| FaQ1fS | Scnd. Code |
|______________________|
<?php
if ($_POST['pass'] == "A5Sr2A") {
{
echo "First Code";
}
} else {
header('Location:index.html');
}
?>
我正在尝试7个小时。真的。我想这样做。我想学习MySQL。请帮助。谢谢
<? ob_start(); ?>
<html>
<link rel=stylesheet href="style.css">
<table>
<tr>
<td>
<center>
<?php
$host = "localhost";
$user = "user";
$password = "pass";
$database = "db";
$con = mysqli_connect($host, $user, $password, $database);
if (mysqli_connect_errno()) {
echo "ERROR";
}
$result = m ysqli_query($con, "SELECT sifre FROM keyler");
if ($_POST['pass'] == $ result) {
{
$mykey = m ysqli_query($con, "SELECT bizimkey FROM keyler");
echo $mykey;
}
} else {
header('Location:index.html');
}
mysqli_close($con);
?>
</center>
</td>
</tr>
</table>
</html>
<? ob_flush(); ?>
好的，请参见下面的基本实现示例：
$host = "localhost";
$user = "user";
$password = "pass";
$database = "db";
$con=mysqli_connect($host,$user,$password,$database);
if(mysqli_connect_errno()!=0)//mysqli_connect_errno() returns 0 when there are no errors
{
echo "ERROR";
}
$sifre_test = mysqli_real_escape_string($con,$_POST['pass']);
$result = mysqli_query($con,"SELECT '1' FROM `keyler` WHERE `sifre`='".$sifre_test."'");
if($result!==false)
{
//If $result !== false, the query was successful, so we'll try to grab a row
$row = mysqli_fetch_assoc($result);
//$row will be null if there wasn't a row found where `sifre`=$sifre_test
if(!is_null($row))
{
$result = mysqli_query($con,"SELECT `bizimkey` FROM `keyler` WHERE `sifre`='".$sifre_test."'");
if($result!==false)
{
$row = mysqli_fetch_assoc($result);
$bizimkey = $row['bizimkey'];
echo $bizimkey;
}
else
{
die("bizimkey query failed");
}
}
}
else
{
die("sifre query failed");
}
你的mysql代码是什么样子的？你有多远？你似乎误解了什么是
mysqli\u queryWHEREsifre$\u POST['pass']