llw*_*llw 7 php mysql variables if-statement conditional-statements
我试图在数据插入MySQL查询之前进行检查.这是代码;
$userid = ($vbulletin->userinfo['userid']);
$sql3 = mysql_query("SELECT * FROM table WHERE ID='$_POST[hiddenID]'");
while ($row = mysql_fetch_array($sql3)){
$toon = $row['toonname'];
$laff = $row['tlaff'];
$type = $row['ttype'];
if ($type == 1){
$type == "Bear";
} elseif ($type == 2){
$type == "Cat";
} elseif ($type == 3){
$type == "Dog";
}
}
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但是,这不起作用.基本上,每种类型的"表格"中都有不同的值.1表示熊,2表示猫,3表示狗.
感谢任何人可以帮助我在脚本中查看问题!
Jez*_*mas 17
你正在比较,而不是分配:
if ($type == 1){
$type = "Bear";
}
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您将值与==或进行比较===.
您可以使用分配值=.
您可以编写更少的代码来实现相同的结果,使用switch语句,或只是一堆if没有elseifs的s.
if ($type == 1) $type = "Bear";
if ($type == 2) $type = "Cat";
if ($type == 3) $type = "Dog";
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我会为它做一个函数,像这样:
function get_species($type) {
switch ($type):
case 1: return 'Bear';
case 2: return 'Cat';
case 3: return 'Dog';
default: return 'Jeff Atwood';
endswitch;
}
$type = get_species($row['ttype']);
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