使jQuery UI自动完成源依赖于另一个自动完成


Make jQuery UI autocomplete source dependent on another autocomplete

问题出在标题上,我不知道如何使自动完成小部件依赖于其他小部件。在我的例子中,我的数据库中有两列:

  • firstName
  • 姓氏

还有一些行:

Dupont Francois
Dupont Oliver
Lapeche Jean 

如果我在第一个自动完成输入中输入Dupont,我希望第二个自动完成建议OliverFrancois而不是Jean

这是我的代码HTML:

<form action="../../controler/add/addUserDevice.php" onsubmit="return verifUserDevice(this)" method="post">
<table>
    <tr>
        <td>Last Name</td> 
        <td><input type="text" class="userLastName" name="userLastName"/></td>
    </tr>
    <tr>
        <td>First Name</td> 
        <td><input type="text" class="userFirstName" name="userFirstName" /</td>    
    </tr>
    <tr>
        <td colspan="3"><input class="buttonIndex" type="submit" value="Create" /></td>
    </tr>
</table>

我的第二个自动完成的PHP代码

<?php
include('../../model/connexion_sql.php');
if ((include_once '../../model/connexion_sql.php') === FALSE)
    exit('erreur include');
if (isset($_GET['term'])) {
    $connection = new Connection;
    $connection->connection();
    $conn    = $connection->getConnection();
    $requete = $conn->prepare('SELECT userFirstName FROM user WHERE userFirstName LIKE :term');
    $requete->execute(array('term' => $_GET['term'] . '%'));
    $array = array();
    while ($donnee = $requete->fetch()) {
        $array[] = $donnee['userFirstName'];
    }
    echo json_encode($array);
}
?>

和我的JavaScript:

$(function(){ 
    $(".userFirstName").autocomplete({
        source: '../../model/autocomplete/userFirstName.php',
        minLength: 1
    });
});
$(function() {   
    $(".userFirstName").autocomplete({   
        autoFocus:true,
        source: function( request, response ) {        
            $.ajax({       
                url: '../../model/autocomplete/userFirstName.php?userLastName=' + $("#userLastName").val() + "&term=" + request.term,   
            dataType: "json",
            success: function( data ) {           
                response(data);      
            }
           });   
        },    
        minLength: 1   
    });
});

您的ajax请求将是这样的。

id="userLastName"添加到您的userLastName输入中。

将此添加到您的PHP脚本中

if($_GET['userLastName']) {
    $userLastName = $_GET['userLastName'];
} else $userLastName = null;

然后更改

SELECT userFirstName FROM user 
WHERE userFirstName LIKE :term

SELECT userFirstName FROM user 
WHERE userFirstName LIKE :term
AND (userLastName = :userLastName OR :userLastName is null)
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