Question

需要一个忙。我想让内部PHP脚本在页面加载时自动运行。我遇到的问题涉及将日期和用户输入的值从表单传递到内部PHP脚本。我尝试了$_POST和$_GET。我应该将脚本放在外部脚本中吗？我真的不需要代码，只要告诉我应该调查什么。谢谢！
输入日志/编辑

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<script type="text/javascript" type="text/javascript" src="js/simpledtpicker.js"></script>
</head>

<body style="text-align:center;">
<div class="wrapper">
<?php include('includes/header.php')?>
<h1 style="text-align:center;">Entry Log/Edit</h1>
<span>
<form action="#" method="GET">
<table style="width:100%;">
<tr>
    <th style="text-align:left;">
    Entries for: 
    <input type="text" name="dateOfEntry" class="dateOfEntry">
    <script type="text/javascript">
            $(function()
            {
                $('*[name=dateOfEntry]').appendDtpicker(
                {
                    "dateOnly": true,
                    "dateFormat": "MM/DD/YYYY",
                    "autodateOnStart": true,
                    "closeOnSelected": true
                });
            });
    </script>         
    </th>
    <th style="text-align:right; padding-right: 0px;">
    User: 
    <input type="text" name="user" value="Hrobi7198228" readonly="readonly" style="border:none; background-color:transparent"/>
    </th>
    <th width="5px;"><input type="submit" name="submit" value="Go"></th>
</tr>
</table>
</form>
</span>
<br>
<table class="entry" style="width: 100%;">
<thead>

<?php 
include ('connServer.php');

    $date = date('Y-m-d', strtotime(str_replace("-","/",$_GET['dateOfEntry'])));
    $username = $_GET['user'];

    $query = 'SELECT `ID`, `Date`, `Description`, `TypeOfDowntime`, `Machine#` FROM `machineissuesreport` WHERE `Date`="'.$date.'" AND `UpdatedBy` = "'.$username.'" ORDER BY `ID` DESC';

    $conn = mysqli_query($connection, $query);

    if ($conn)
    {
        echo '<tr>
              <th width="5px">Edit</th>
              <th width="5px">Delete</th>
              <th>Date</th>
              <th>Details</th>
              <th width="100px" style="text-align:center;">Downtime Type</th>
              <th width="20px" style="text-align:center;">Machine No.</th>
              </tr>
              </thead>
              <tbody>';
        while($row = mysqli_fetch_array($conn))
        {
            echo '<tr>';
            echo '<td style="text-align: center" width="5px"><a href="editMachineIssue.php?id='.$row["ID"].'" class="edit">Edit</a></td>';
            echo '<td style="text-align: center" width="5px"><a href="#" id="'.$row['ID'].'" class="delete">Delete</a></td>';
            echo '<td style="display: none;"><input type="hidden" value='.$row['ID'].'></td>';
            echo '<td width="5px" style="text-align: center; padding: 0px 5px;">'.$row['Date'].'</td>';
            echo '<td style="text-align: left; padding: 0px 0px 0px 12px;">'.$row['Description'].'</td>';
            echo '<td style="text-align: center;">'.$row['TypeOfDowntime'].'</td>';
            echo '<td style="text-align: center;">'.$row['Machine#'].'</td>';
            echo '</tr>';
        }
    }


?>

<script type="text/javascript">
$(document).ready(function() 
{
    $.ajax
    ({
        type: 'POST', 
        data: {'run':run},
        success: function(data)
        {
        }
    });

    $('.delete').click(function()
    {
        if(confirm("Are you sure you want to delete this row?"))
        {
            var del_id = $(this).attr('id');
            var $ele = $(this).parent().parent();

            $.ajax({
                type: 'POST',
                url: 'machineEntryLogEdit.php',
                data: {'del_id' : del_id},
                success: function(data)
                {
                    $ele.fadeOut().remove();            
                },
                error: function (xhr, status, error)
                {
                    alert(this);
                }
            });
        }
    });
});
</script>
</tbody>
</table>
</div>
</body>
</html>

我仍然在努力，但请帮助我找到适合Google的条款。我是PHP和AJAX的新手。

Answer 1

您可以在js文件中编写以下代码，并在页面中包含js文件。

$(document).ready(function(){
$.ajax({ url: "file.php",
    data: data1,
    success: function(){
        // Do whatever you want to do here.
    }});
});

PHP脚本在页面加载时自动运行

1 个答案: