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我有一个显示 mysql 记录的下拉选择。我想要的是创建另一个下拉选择以拥有一个组合框。

出价.php

<script type="text/javascript" src="javascript/jquery.js"></script>
<script>
function showUser(str) {
    var $txtHint = $('#txtHint');
    if (str == "") {
        $txtHint.html('');
        return;
    }
    $txtHint.load('bid_list.php?q=' + str)
}
</script>
</head>
<body onload=showUser(str="ALL")>

<div id="main" style="width:1150px;margin:0 auto;padding:10px;">
<div id="title" style="width:1150px;margin:0 auto;text-align:center;font-size:30px;color:gray;text-shadow:2px 2px #000;font-family:Verdana, Tahoma, sans-serif;"><h3>BIDDING LIST</h3></div>
<div id="container" style="width:1100px;">
<?php
$mysqli = new mysqli("localhost", "root", "", "app");
$result = $mysqli->query("SELECT bid FROM procurement WHERE bid LIKE '13-___' OR bid LIKE '1_-___' OR bid LIKE '2_-___' GROUP BY bid ORDER BY bid");

$option = '';
while($row = $result->fetch_assoc())
{
  $option .= '<option value = "'.$row['bid'].'">'.$row['bid'].'</option>';
}
?>

<div style="text-align:center;">
<select name="users" onchange="showUser(this.value)" style="overflow:scroll;width:100px;">
        <option value="ALL" selected='ALL'>ALL</option>
        <?php echo $option; ?>
</select>
</div>
<br>

<div id="txtHint"></div>

出价清单.php

<html>
    <head>
        <title></title>
        <link rel="stylesheet" href="css/style.css" type="text/css" id="" media="print, projection, screen" />
        <script type="text/javascript" src="javascript/jquery.tablesorter.js"></script>
        <script type="text/javascript">
        $(function() {
            $("table").tablesorter({debug: true});
        });
        </script>
<script type="text/javascript">
    $(function(){
    var tfrow = document.getElementById('tfhover').rows.length;
    var tbRow=[];
    for (var i=1;i<tfrow;i++) {
        tbRow[i]=document.getElementById('tfhover').rows[i];
        tbRow[i].onmouseover = function(){
          this.style.backgroundColor = '#f3f8aa';
        };
        tbRow[i].onmouseout = function() {
          this.style.backgroundColor = '#ffffff';
        };
    }
    });
</script>
</head>
<body>

<?php
$mysqli = new mysqli("localhost", "root", "", "app");
$q = $_GET["q"];
$where = '';
if ( $q != 'ALL' ) {
    $where = " WHERE bid='$q' ";
}
$result1 = $mysqli->query("
    SELECT bid, item_name, item_description, unit, unit_cost, quantity, supplier, po_number, po_date, counter, SUM(unit_cost*quantity) AS total_amount 
    FROM procurement 
    $where 
    GROUP BY counter ORDER BY bid
");
echo'<table id="tfhover" cellspacing="0" class="tablesorter" style="text-transform:uppercase;">
        <thead>
        <tr>
            <th title="Item Name">Item Name</th>
            <th title="Item Description">Description</th>
            <th title="Example : Pc, Pcs, Box and Etc.">Unit</th>
            <th title="Item Price">Unit Cost</th>
            <th title="Total Item Quantity">QTY</th>
            <th title="Total Price">Total Amount</th>
            <th title="Name of Supplier">Supplier</th>
            <th title="Purchase Order #">PO #</th>
            <th title="Purchase Order Date">PO Date</th>
        </tr>
        </thead>';
        echo'<tbody>';
while($row = $result1->fetch_assoc()){
if($row['bid'] != '')
 {
 echo'<tr>
            <td>'.$row['item_name'].'</td>
            <td>'.$row['item_description'].'</td>
            <td>'.$row['unit'].'</td>
            <td>'.number_format($row['unit_cost'], 2, '.', ',').'</td>
            <td>'.$row['quantity'].'</td>
            <td>'.number_format($row['total_amount'], 2, '.', ',').'</td>
            <td>'.$row['supplier'].'</td>
            <td>'.$row['po_number'].'</td>
            <td>'.$row['po_date'].'</td>
       </tr></tbody>';
       }
        }
    echo "</table>";

if (!$mysqli) {
    die('Connect Error: ' . mysqli_connect_error());
}
mysqli_close($mysqli);
    ?> 

</body>
</html>

如果下拉值等于 13-001，我想要做什么，第二个下拉列表显示所有 po_number，如果我在第二个下拉列表中选择 po_number 的所有记录，如表中显示的 12-12-CO 397。怎么做？

4

1 回答 1

0

这一切都是我自己想出来的，希望对你有所帮助。

在视图中：

<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="/js/dynamicSelect.js"></script>
<script>
$( document ).ready(function() {
    $("#selectToUpdate").dynamicSelect($("#changeTriggersUpdate"),"actionName.php");
});
</script>

我做了一个 jQuery 插件 dynamicSelect.js '$.fn.x =' 将它添加到 jQuery

$.fn.dynamicSelect = function(trigger, action) {
    var dynamicSelect   = $(this);
    trigger.on('change', function() {
        if(trigger.val() > 0){   
            $.getJSON(action, {searchVal: $(this).val()}, function(html) {
                dynamicSelect.html(html);
            });
        } else{
            dynamicSelect.html('');
        }
    });
};

我为 Yii 框架编写了我的服务器端代码，所以它不会对你有多大好处，但它应该让你知道需要做什么。

    public function actionVehicleSearch()
{
            $searchVal = $_REQUEST['searchVal'];
            $vehicles = Vehicle::model()->findAll("
                       customerID LIKE  ?
                        ", array('%' . $searchVal . '%'));
            $str = '';
            foreach ($vehicles as $vehicle) {
                $str .= '<option value="'. $vehicle->vID .'">'. $vehicle->VIN .'</option>';
            }

            echo CJSON::encode($str);
}
于 2013-11-17T08:03:14.053 回答