在城市下拉列表中保留价值（第 4 部分）答案

【问题标题】：keep value in city drop down list(part 4)在城市下拉列表中保留价值（第 4 部分）
【发布时间】：2014-11-03 08:07:50
【问题描述】：

续。来自Get state name from drop down list after click submit button (part 3)

状态数据 json ($stateJsonObject)

Array ( 
[0] => stdClass Object ( [stateId] => s1 [stateName] => Kuala Lumpur) 
[1] => stdClass Object ( [stateId] => s2 [stateName] => Selangor)
)

城市数据 json ($cityJsonObject)

Array ( 
[0] => stdClass Object 
( [cityId] => c1 [cityName] => Kajang [cityStateId] => s2 ) 
[1] => stdClass Object 
( [cityId] => c2 [cityName] => Seputeh [cityStateId] => s1 ) 
[2] => stdClass Object ( [cityId] => c3 [cityName] => Shah Alam [cityStateId] => 
s2 ) 
[3] => stdClass Object ( [cityId] => c4 [cityName] => Klang [cityStateId] => s2  
) 
[4] => stdClass Object ( [cityId] => c5 [cityName] => Kepong [cityStateId] => s1    
)
)

代码（test3.php）

<?php
    $cityState = array();
    $cityName = array();

    for($j = 0; $j < count($cityJsonObject); $j++)
    {
        $cityState[] = $cityJsonObject[$j] -> cityStateId;
        $cityName[] = $cityJsonObject[$j] -> cityName;
    }
?>

<html>
<head>
<script type="text/javascript">
    function showCity(state, target_id)
    {
        var stateId = state.options[state.selectedIndex].value; 

        var target = document.getElementById(target_id);
        target.length = 0;
        target.options[0] = new Option('select one', '');
        target.selectedIndex = 0;

        var cityState = <?php echo json_encode($cityState, JSON_HEX_QUOT) ?>;
        var cityName = <?php echo json_encode($cityName, JSON_HEX_QUOT)?>; 

        for(k = 0; k < cityState.length; k++)
        {
            if(stateId == cityState[k])
            {
                target.options[target.length] =
                new Option(cityName[k],cityName[k]);         
            }
        }  
    }
</script>
</head>

<body>
<form action="test3.php" method="post">
    State:
    <select name="state" id="state" onchange="showCity(this, 'city')">
        <option value ="">select one</option>
        <?php
            $select_sign = '';
            for($i = 0; $i < count($stateJsonObject); $i++)
            {
                if($stateJsonObject[$i] -> stateId == $_POST['state']) 
                {$select_sign = "SELECTED";}else{$select_sign = "";}
                echo '<option value = '.$stateJsonObject[$i] -> stateId.' 
                '.$select_sign.'>';
                echo $stateJsonObject[$i] -> stateName;
                echo '</option>';
            }
        ?>
    </select>

    <br />

    City:
    <select name="city" id="city">
        <option value ="">select one</option>
    </select>

    <br />

    <input type="submit" name="submit" value="Submit"/>
</form>        
</body>
</html>

我的问题是我从州下拉列表中选择雪兰莪，然后我从城市下拉列表中选择巴生。点击提交按钮后，我应该如何在城市下拉列表中保持巴生名称被选中？

【问题讨论】：

标签： javascript php json

【解决方案1】：

您可以使用以下任一选项，

使用 AJAX 提交表单。然后页面将不会重新加载并且您可以保持该值不变。
使用 HTML5 本地存储（或 cookie）来跟踪城市名称应该选择的。页面加载时，检查城市的值是否已保存，如果已保存，请将选择设置为已保存的值。

【讨论】：