Php 动态创建mysql搜索字符串?

Php 动态创建mysql搜索字符串?,php,mysql,search,Php,Mysql,Search,我正在尝试创建一个简单的搜索页面,但我不能100%确定如何使用适当的AND等来编写实际的搜索字符串如果变量存在,代码如下: if ($post) { //get all search variables $type = JRequest::getVar('type'); $classifications = JRequest::getVar('classifications', array(0), 'post', 'array'); $rating = JReq

我正在尝试创建一个简单的搜索页面,但我不能100%确定如何使用适当的AND等来编写实际的搜索字符串如果变量存在,代码如下:

if ($post) {

    //get all search variables
    $type = JRequest::getVar('type');
    $classifications = JRequest::getVar('classifications', array(0), 'post', 'array');
    $rating = JRequest::getVar('rating');
    $status = JRequest::getVar('status');
    $cterms = JRequest::getVar('cterms');
    $clientid = JRequest::getVar('clientid');
    $company = JRequest::getVar('company');
    $address = JRequest::getVar('address');
    $name = JRequest::getVar('name');
    $surname = JRequest::getVar('surname');
    $city = JRequest::getVar('city');
    $state = JRequest::getVar('state');
    $pcode = JRequest::getVar('pcode');
    $country = JRequest::getVar('country');

    //create search string
    echo "SELECT * FROM #__db_clients "; <- the query is supposed to be done here.. it's in as echo because I was trying to spit it out before trying to make it run.. :)

} else {

    echo 'There has been an error, please try again.';

};

我尝试过使用if type!=null then searchtype=where type='X',但如果搜索需要,我无法确定如何放置前/后。。这有意义吗

这是一个简单的例子。我不知道什么样的数据JRequest::getVar总是返回字符串或混合类型?但这应该让你开始。确保在foreach循环中使用任何适用的转义方法:

if ($post) {
    $criteria = array();
    //get all search variables
    $criteria['type'] = JRequest::getVar('type');
    $criteria['classifications'] = JRequest::getVar('classifications', array(0), 'post', 'array');
    $criteria['rating'] = JRequest::getVar('rating');

    //if there are some criteria, make an array of fieldName=>Value maps
    if(!empty($criteria)) {
        $where = array();
        foreach($criteria as $k => $v) {
            //IMPORTANT!!
            //$v is the value of the field, needs to be quoted correctly!!
            $where[] = "$k = '$v'";
        }
    }
    //create search string
    $query =  "SELECT * FROM #__db_clients";

    if($where) {
        $query .= " where " . join(' AND ', $where);
    }   
} else {    
    echo 'There has been an error, please try again.';
};

在表达式中插入$v时,一定要将转义应用于它。@Bill Karwin-谢谢,绝对感谢!我不知道哪种转义方法适用于他的应用程序,所以我在答案和代码注释中都添加了注释。非常感谢,这肯定会帮助我运行代码:非常感谢@SoulieBaby-np,确保彻底测试你将其转化为什么,它有助于在每个步骤中放置大量诊断回音语句,以便你可以看到发生了什么。并确保所有内容都正确转义/引用。祝你好运嗯,它不喜欢foreach$条件,因为$k=$v{我得到这个错误:解析错误:语法错误,意外的'=',期望