如何插入值从下拉到mysql查询检索数据从数据库 - how to insert value from drop down into mysql query to retrive data from data base

how to insert value from drop down into mysql query to retrive data from data base

我必须从下拉表中选择城市，根据值，我必须使用jquery进一步检索属于该城市的人我的代码是

 <select class="form-control" id="personCity">
            <option class="form-control" value="">Person country</option>
            <option class="form-control" value="usa">usa</option>
            <option class="form-control" value="aus">Australia/option>
</select>

我必须得到使用jquery的值

$('#personCity').change(function(e) {
   var personCity=$('#personCity').val();
   $('#personcityRetrive').text(personCity);
});

和在数据库中我有以下数据

country     person
 usa           abc
 aus           xyz
 usa           123
 aus           ABC

mysql代码

<?php
$sql=mysql_query("SELECT * FROM outlets WHERE country='" 
?><span id="personcityRetrive"></span> 
<?php "'");?>

给定这个HTML:

<select class="form-control" id="personCity">
    <option class="form-control" value="">Person country</option>
    <option class="form-control" value="usa">usa</option>
    <option class="form-control" value="aus">Australia/option>
</select>

您需要在发生更改时执行AJAX。JQuery:

$('#personCity').change(function(e) {
    var personCity=$('#personCity').val();
    $.get('lookupCity.php', { "city": personCity }, function(result){
        $("#personcityRetrive").html(result);
    });
});

您的PHP将需要独立(lookupCity.php)，连接到SQL Server，并回显SQL查找的结果。它不应该使用MySQL函数，这些已被弃用。您还需要避免SQL注入。试试这个:

<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s'n", mysqli_connect_error());
    exit();
}
if ($stmt = mysqli_prepare($link, "SELECT person FROM outlets WHERE country=? LIMIT 1");
    mysqli_stmt_bind_param($stmt, "s", $_GET['city']);
    mysqli_stmt_execute($stmt);
    mysqli_stmt_bind_result($stmt, $person);
    mysqli_stmt_fetch($stmt);
    printf("%s'n", $person);
    mysqli_stmt_close($stmt);
}
mysqli_close($link);
?>