1

有人请帮忙!这可能是我犯的一个非常简单的错误,但我是 PHP 和 MySQL 的新手。我正在尝试根据用户的日期选择从数据库中检索信息。我正在使用我在 youtube 上找到的一些示例将输入发布到带有 jquery 的 php 页面,并在 html 页面上显示从 php 检索到的值。但是,它似乎只返回下面的 else if 语句。非常感谢您的帮助!

下面的HTML:

<div id="customDateRange">
    <h2>Welcome <?php echo $companyName ?></h2>
    <h3 style="color:#ff0000">This is currently under development</h3>
    <h4>Please specify the dates you would like to view below:</h4>
    <label for="dateStart">Start Date</label>
    <input class="datePick" id="dateStart" name="dateStart" type="text"/>
    <label for="dateEnd">End Date</label>
    <input class="datePick" ud="dateEnd" type="text"/>
    <input id="getDateRange" type="submit" value="Display Report" />
    <div id="fromDatabase"></div>
</div>

下面的Javascript:

$('.datePick').datepicker({ minDate: new Date(2013, 3 - 1, 1), dateFormat: 'yy-mm-dd' });

$('input#getDateRange').on('click', function(){
    var startDate = $('input#dateStart').val();
    if ($.trim(startDate) != '') {
        $.post('../../includes/dateRangeDB.php', function(data, status){
            $('div#fromDatabase').text(data);
        });
    }
});

PHP 下面:

<?php 
if (isset($_POST['dateStart']) === true && empty($_post['dateStart']) === false) {
    $query = mysql_query("
        SELECT TotalCost FROM adwords_reporting WHERE CampaignID = 'CAMP020' AND Date IN ('" . mysql_real_escape_string(trim($_POST['dateStart'])) . "')
    ");

    echo "Total Cost: " . (mysql_num_rows($query) !== 0) ? mysql_result($query, 0, 'TotalCost') :  'Date Not Found';
}
else if (isset($_POST['dateStart']) === false) {
    echo "WTF MAN!";
}
else {
    echo "It seems there is an issue";
}
?>

感谢您的帮助!

4

1 回答 1

3

您需要实际发送日期:

$('input#getDateRange').on('click', function(){
    var startDate = $('input#dateStart').val();
    if ($.trim(startDate) != '') {
        $.post('../../includes/dateRangeDB.php', {dateStart: startDate}, function(data){
            $('div#fromDatabase').text(data);
        });
    }
});

能够访问它:

$_POST['dateStart']
于 2013-03-22T21:04:43.597 回答