Question

我正在尝试使用php和mysql创建一个简单的搜索脚本。我有html选择标签

人
国家
区域
目的地
来自
到

有了这个，我从mysql数据库中获取内容。以下是我的PHP脚本。

if(isset($_GET['Submit']) && $_GET['Submit'] == "Search")
{
$people = mysql_real_escape_string(htmlspecialchars(trim($_GET['people'])));
$country =  mysql_real_escape_string(htmlspecialchars(trim($_GET['country'])));
$region =  mysql_real_escape_string(htmlspecialchars(trim($_GET['region-depart'])));
$destination = mysql_real_escape_string(htmlspecialchars(trim($_GET['destination'])));
$from = mysql_real_escape_string(htmlspecialchars(trim($_GET['from'])));
$to = mysql_real_escape_string(htmlspecialchars(trim($_GET['to'])));

if(isset($people))
{

$search = mysql_query("SELECT * FROM property_step1 WHERE pro_no_sleep LIKE 
'%$people%'");
$num = mysql_num_rows($search);

while($result = mysql_fetch_array($search))
    {
        $propertyid = (int) $result['propertyid'];          
        echo $country_d = $result['pro_country'];
        echo $region_d = $result['pro_state'];
        echo $destination_d = $result['pro_city'];

    }
}

elseif(isset($country))
{
$search2 = mysql_query("SELECT * FROM property_step1 WHERE pro_country LIKE 
'%$country%'");
$num = mysql_num_rows($search2);        

while($result2 = mysql_fetch_array($search2))
    {
        $propertyid = (int) $result2['propertyid'];         
        echo $country_d = $result2['pro_country'];
        echo $region_d = $result2['pro_state'];
        echo $destination_d = $result2['pro_city'];

    }
}
else
{
    echo "nope";
}       
}

好吧，如果我选择人（值为1,2,3等），当我选择国家/地区时，它会显示数据库但的内容它没有显示任何内容。我的查询有什么问题吗？

Answer 1

isset($people)始终评估为true;你需要检查它是否也是empty：

if (isset($people) && !empty($people)) {
    // ...
}

Answer 2

国家/地区的 elseif 条件正在创建问题仅用 if 替换它，写if...elseif只有一个条件会被执行。

使用此代码

if (isset($_GET['Submit']) && $_GET['Submit'] == "Search") {
    $people = mysql_real_escape_string(htmlspecialchars(trim($_GET['people'])));
    $country = mysql_real_escape_string(htmlspecialchars(trim($_GET['country'])));
    $region = mysql_real_escape_string(htmlspecialchars(trim($_GET['region-depart'])));
    $destination = mysql_real_escape_string(htmlspecialchars(trim($_GET['destination'])));
    $from = mysql_real_escape_string(htmlspecialchars(trim($_GET['from'])));
    $to = mysql_real_escape_string(htmlspecialchars(trim($_GET['to'])));

    if (isset($people)) {
        $search = mysql_query("SELECT * FROM property_step1 WHERE pro_no_sleep LIKE 
'%$people%'");
        $num = mysql_num_rows($search);

        while ($result = mysql_fetch_array($search)) {
            $propertyid = (int) $result['propertyid'];
            echo $country_d = $result['pro_country'];
            echo $region_d = $result['pro_state'];
            echo $destination_d = $result['pro_city'];
        }
    } 
    if (isset($country)) {
        $search2 = mysql_query("SELECT * FROM property_step1 WHERE pro_country LIKE 
'%$country%'");
        $num = mysql_num_rows($search2);

        while ($result2 = mysql_fetch_array($search2)) {
            $propertyid = (int) $result2['propertyid'];
            echo $country_d = $result2['pro_country'];
            echo $region_d = $result2['pro_state'];
            echo $destination_d = $result2['pro_city'];
        }
    } else {
        echo "nope";
    }
}

Answer 3

您正在定义每个变量，因此所有变量将始终“设置”。

if(isset($people))将始终运行，因为它的定义意味着isset($country)永远不会运行。

这需要更改为：

if(!empty($people)){

}
if(!empty($country)){

}

Php Mysql搜索问题

3 个答案: