Question

我一直在想办法如何提取当前页面元素和内容以便可以使用它，我可以将PDF作为输出。目前我的reports2.php包含：

<body style="padding: 10px; font-size: 10pt;" onLoad="generateReport()">
        <hr>
        <div backcolor="#FEFEFE" backimg="./res/bas_page.png" backimgx="center" backimgy="bottom" backimgw="100%" backtop="0" backbottom="30mm" footer="date;heure;page" style="font-size: 12pt">
            <hr>
            <table cellspacing="0" style="width: 100%; text-align: center; font-size: 14px">
                <tbody>
                    <tr>
                        <td style="width: 75%;">
                            <h1> Advanced World Solutions, Inc.</h1>
                            <h2> Actual Physical Inventory of MIS Hardware Resources </h2>
                        </td>
                        <td style="width: 25%; color: #444444;">
                            <img style="width: 100%;" src="C:\wamp\www\test\images\logo.jpg"><br>
                        </td>
                    </tr>
                </tbody>
            </table>
            <hr>
            <table id="search" border="1" style="width:100%">
            </table>
        </div>
        <hr>
    </body>

我想要的是在运行此脚本后，页面内容和脚本输出将创建一个新的.php或.html文件，然后我将其转换为PDF。

function generateReport()
{   
    var xhr;  
    if (window.XMLHttpRequest) {
            xhr = new XMLHttpRequest();  
    } 

    else if (window.ActiveXObject) {
            xhr = new ActiveXObject("Microsoft.XMLHTTP");
    }  
    var data = "asid=0"
    xhr.open("POST", "report/genrep.php", true);   
        xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");              
        xhr.send(data);  
        xhr.onreadystatechange = display_data; 
        function display_data() 
    {  
            if (xhr.readyState == 4) 
        { 
                if (xhr.status == 200) 
            {       
                    document.getElementById("search").innerHTML = xhr.responseText;
                } 
            else 
            { 
                alert('There was a problem with the request.');  
                }  
            }  
        }  
}

我尝试过使用

file_put_contents('test.php', ob_get_contents());

但它返回没有脚本输出的页面。

Answer 1

也许这会帮助您获取页面内容（仅限于正文内部）..：D

function produce(){
	var data = "<textarea rows='50' cols='50' style='width:100%;'>"+document.body.innerHTML+"</textarea>";
	document.body.innerHTML = data+document.body.innerHTML;
}

<!-- Any contents as you like -->
..
..

<button id="btn" onclick="produce();">Click to produce result</button>

运行脚本后获取当前页面内容

1 个答案: