为什么我的PHP MYSQL比较不起作用_Php_Mysql_If Statement_Compare

为什么我的PHP MYSQL比较不起作用

php mysql if-statement

为什么我的PHP MYSQL比较不起作用,php,mysql,if-statement,compare,Php,Mysql,If Statement,Compare,我在胡闹，想看看我是否能制作一个几年前风靡一时的可点击宠物网站，但我在尝试使用PHP中的if，elseif东西时遇到了一个问题。以下是我所拥有的： <?php include_once "mysql_connect.php"; $newip = $_SERVER['REMOTE_ADDR']; $oldip = mysql_query("SELECT lastip FROM sitefunctions WHERE name='index'"); if ($newip == $oldi

我在胡闹，想看看我是否能制作一个几年前风靡一时的可点击宠物网站，但我在尝试使用PHP中的if，elseif东西时遇到了一个问题。以下是我所拥有的：

<?php
include_once "mysql_connect.php";

$newip = $_SERVER['REMOTE_ADDR'];
$oldip = mysql_query("SELECT lastip FROM sitefunctions WHERE name='index'");

if ($newip == $oldip) {
$message = "You were the last one to click this pet, please wait until someone else has clicked it before trying again.";
}
else {
mysql_query("UPDATE sitefunctions SET `clicks` = `clicks`+1 WHERE name='index'");
mysql_query("UPDATE sitefunctions SET `lastip` = '$newip'  WHERE name='index'");
$tempclicks = mysql_query("SELECT `clicks` FROM sitefunctions WHERE name='index'");
$message = "You gave this pet a click!";
};
if ($tempclicks == 150) {
mysql_query("UPDATE sitefunctions SET `level` = 2 WHERE name='index'");
$message = "Your click leveled the pet up!";
}
elseif ($tempclicks == 600) {
mysql_query("UPDATE sitefunctions SET `level` = 3 WHERE name='index'");
$message = "Your click leveled the pet up!";
}
$sql = mysql_query("SELECT * FROM sitefunctions WHERE name='index'");
while($row = mysql_fetch_array($sql)){
$clicks = $row["clicks"];
$level = $row["level"];
$name = $row["name"];
$image1 = $row["image1"];
$image2 = $row["image2"];
$image3 = $row["image3"];
};
if ($level == 1) {
$imageu = $image1;
}
elseif ($level == 2) {
$imageu = $image2;
}
elseif ($level == 3) {
$imageu = $image3;
}
?>
<html>
<head>
</head>
<body>
<p>
<?php print $oldip; ?> <br>
<?php print $newip; ?> <br>
Name: <?php print $name; ?> <br>
<img src=<?php print $imageu; ?> /> <br>
Clicks: <?php print $clicks; ?> <br>
Level: <?php print $level; ?> <br>
<?php print $message; ?>
</p>
</body>
</html>

无论我尝试了什么，它都不能真正比较这些值。如果我写一个“=”表示无论发生什么，它们都是一样的，如果我写“=”则表示它们是不同的，即使它们不应该是不同的

我甚至不知道从哪里开始，没有错误出现，而且我对PHP和MYSQL相当陌生。在这之前，没有其他东西可以真正测试，但我确信其余的比较也不起作用

我在我的网站上使用000webhost，如果已知有问题，lol

这就是我的代码现在的样子，它也可以工作，所以我在这里完成了：

<?php error_reporting(E_ALL); ini_set('display_errors', 1);
$name = $_POST['name'];
if (empty($name) == true){
$name = "index";
};

include_once "mysql_connect.php";
$newip = $_SERVER['REMOTE_ADDR'];

$sql = mysql_query("SELECT * FROM sitefunctions WHERE name='$name'") or die(mysql_error());
while($row = mysql_fetch_array($sql)) {
$lastip = $row["lastip"];
}
if ($lastip == $newip) {
$message = "You were the last one to click this pet! You have to wait until someone else clicks it!";
} else {
mysql_query("UPDATE sitefunctions SET `clicks` = `clicks`+1 WHERE name='$name'") or die(mysql_error());
mysql_query("UPDATE sitefunctions SET `lastip` = '$newip' WHERE name='$name'") or die(mysql_error());
$message = "You clicked the pet!";
}
$sql = mysql_query("SELECT * FROM sitefunctions WHERE name='$name'") or die(mysql_error());
while($row = mysql_fetch_array($sql)) {
$clicks = $row["clicks"];
$level = $row["level"];
}

if ($clicks > 50*$level) {
mysql_query("UPDATE sitefunctions SET `level` = `level`+1 WHERE name='$name'") or die(mysql_error());
$message = "Your click leveled up the pet!";
}

$sql = mysql_query("SELECT * FROM sitefunctions WHERE name='$name'") or die(mysql_error());
while($row = mysql_fetch_array($sql)) {
$clicks = $row["clicks"];
$level = $row["level"];
$name = $row["name"];
$image1 = $row["image1"];
$image2 = $row["image2"];
$image3 = $row["image3"];
$lastip = $row["lastip"];
};
if ($level > 35) {
$imageu = $image3;
} elseif ($level > 15) {
$imageu = $image2;
} elseif ($level > 0) {
$imageu = $image1;
};
?>
<html>
<head>
</head>
<body>
<center>
<p>
Name: <?php print $name; ?> <br>
<img src=<?php print $imageu; ?> /> <br>
Clicks: <?php print $clicks; ?> <br>
Level: <?php print $level; ?> <br>
Last User: <?php print $lastip; ?> <br>
<?php print $message; ?>
</p>
</center>
</body>
</html>

从哪里开始？单个=
是一个赋值，而不是一个比较，因此=
或可能的==
是您需要的。您的比较失败，因为您没有获取查询结果，也没有检查结果是否有错误，而且您使用的是不推荐使用的API。根据mysqli
或PDO
找到一个好的教程，并遵循它。这一点太宽泛了，我倾向于说这段代码似乎是正确的，但我的印象是，您的服务器可能不再支持较旧的mysql\uu
功能。在打开PHP标记后立即将错误报告添加到文件顶部，例如哈哈，谢谢大家的输入，我一直在浏览教程，很难找到我需要的内容，我注意到了所有更新，所以我认为类似的情况可能会发生。至于if（$newip=$oldip）
-其中name='index'必须是该行中唯一的条目。如果有其他相同的，那么这可能是一个问题。另外，000webhost确实支持mysql，因此您的查询可能会在某个地方失败。问题是，是的，它可以抓取信息并打印出来（代码的底部），但不能用于比较。
<?php error_reporting(E_ALL); ini_set('display_errors', 1);
$name = $_POST['name'];
if (empty($name) == true){
$name = "index";
};

include_once "mysql_connect.php";
$newip = $_SERVER['REMOTE_ADDR'];

$sql = mysql_query("SELECT * FROM sitefunctions WHERE name='$name'") or die(mysql_error());
while($row = mysql_fetch_array($sql)) {
$lastip = $row["lastip"];
}
if ($lastip == $newip) {
$message = "You were the last one to click this pet! You have to wait until someone else clicks it!";
} else {
mysql_query("UPDATE sitefunctions SET `clicks` = `clicks`+1 WHERE name='$name'") or die(mysql_error());
mysql_query("UPDATE sitefunctions SET `lastip` = '$newip' WHERE name='$name'") or die(mysql_error());
$message = "You clicked the pet!";
}
$sql = mysql_query("SELECT * FROM sitefunctions WHERE name='$name'") or die(mysql_error());
while($row = mysql_fetch_array($sql)) {
$clicks = $row["clicks"];
$level = $row["level"];
}

if ($clicks > 50*$level) {
mysql_query("UPDATE sitefunctions SET `level` = `level`+1 WHERE name='$name'") or die(mysql_error());
$message = "Your click leveled up the pet!";
}

$sql = mysql_query("SELECT * FROM sitefunctions WHERE name='$name'") or die(mysql_error());
while($row = mysql_fetch_array($sql)) {
$clicks = $row["clicks"];
$level = $row["level"];
$name = $row["name"];
$image1 = $row["image1"];
$image2 = $row["image2"];
$image3 = $row["image3"];
$lastip = $row["lastip"];
};
if ($level > 35) {
$imageu = $image3;
} elseif ($level > 15) {
$imageu = $image2;
} elseif ($level > 0) {
$imageu = $image1;
};
?>
<html>
<head>
</head>
<body>
<center>
<p>
Name: <?php print $name; ?> <br>
<img src=<?php print $imageu; ?> /> <br>
Clicks: <?php print $clicks; ?> <br>
Level: <?php print $level; ?> <br>
Last User: <?php print $lastip; ?> <br>
<?php print $message; ?>
</p>
</center>
</body>
</html>