我有一个页面,根据$_POST变量建立一个非常长的查询.我试图在没有MySQL的情况下这样做的原因OR是因为我必须重新整理整个查询只是为了在最后改变一个小参数,这似乎是浪费资源.
这是我的伪查询:
SELECT * FROM names WHERE
person = '$var1' AND
place = '$var2' AND
location_a = '$var3' AND
location_b = '$var3'
如果我要使用OR,这个胖丑陋的查询将是:
SELECT * FROM names WHERE
person = '$var1' AND
place = '$var2' AND
location_a = '$var3' OR
person = '$var1' AND
place = '$var2' AND
location_b = '$var3'
有没有一种方法可以同时搜索location_a和location_b对$var3不使用OR？
$('#button').live('click', function () {
values_array = [];
$('.field').each(function () {
values_array.push = $(this).val();
});
$.ajax({
url: 'page.php',
data: {
array_a: values_array
},
type: 'POST',
success: function (data) {
$('#div').html(data);
}
});
});
//page.php
echo $_POST['array_a'] . "<br/>"; //The line break echos, but nothing else
答:我是否需要迭代每个类$.each以创建一个正确的数组,和
B.)为什么不回复它？
你如何迭代所有选择的当前值,而不是选项？
我尝试过这种变化:
$('select[name^="mydropdowns"] :selected').each(function() {
if ($(this).val() != 'Test') {
alert("Found.");
}
});
"测试"是两个下拉菜单中的默认选定值,但此脚本从不返回警报.我在这做错了什么？
//DB connect is here
foreach ($find as $listing) {
//bunch of hooblah that discovers $state and $city
//this and below is all you need to see really
$city = strip_tags($location_broken3[0]);
$state = strip_tags($location_broken3[1]);
print $city; //THIS WORKS...
print $state; //AS DOES THIS!
//tried this also
$city1 = settype($city, 'string');
$state1 = settype($state, 'string');
$zip_query = mysql_query("SELECT * FROM zip_codes WHERE city = '$city'
AND state = '$state' OR city = '$city' AND full_state = '$state'
") or die(mysql_error());
if ($row …如果我在我的服务器根目录中存储了一组登录凭据,如果我只是将它存储在PHP文件中,是否存在安全风险？
$user = 'user';
$password = 'password';
我将登录凭据存储到我的网站上具有锁定内容的页面,用户必须单击要接收的链接,这是您想知道的.当用户登录时,我可以像这样授权登录,对吧？
include('../logininfo.php');
if ($_POST['user'] == $user) {
//...etc
这种方法有问题吗？