contenteditable="true" cannot be submitted to php, what should I do?

Question

Recently I am developing a rich text editor. Considering that textarea can only input text, I use the contenteditable="true" attribute of p to implement rich text, which can insert pictures, videos, etc.

But here comes the problem, on the form page:

<form action="test.php" method="post">
<p contenteditable="true" name="zhengwen"></p>
<input type="submit">
</form>

In this way, the content in p cannot be submitted to the test.php page at all, and echo $_POST[zhengwen]; is not output.

I suspect it was not submitted at all.

Does anyone know how to solve it? Looking for simple codes for form submission page and receiving page! ! !

Answer 1

It is indeed not submitted because the submit button in the form only submits the form elements. And <p contenteditable="true"> is not a form element.

If you want to submit the information of <p contenteditable="true">, you need to construct the POST request yourself. I will use jQuery as an example:

$('form').submit(function(e) {
    e.preventDefault(); // 拦截HTML的默认表单提交

    var content = $('p[name=zhengwen]').html();
    $.post('....', {zhengwen: content}, functino(data) {
        // 成功回调
    });
});

Of course, there are many ways to write jQuery's POST submission, such as $.ajax(), etc.

When receiving on the PHP side, just read $_POST as usual.

Answer 2

1、富文本编辑器可以用百度的UEditor
2、正如楼上所说，你可以用ajax来提交，但是这个有一点不好，如果有一百个输入框，难道提交一百个键值对？
3、所以你可以用js的formData对象，图片也可以发送过去，代码如下
    $("#submit").click(function() {
        var x = new FormData(document.getElementById("frm"));//构造方法里面必须是dom对象
        x.append('abc', 123);//追加你的数据
        $.ajax({
            url: '1.php',
            type: 'POST',
            data: x,
            processData: false,  // 告诉jQuery不要去处理发送的数据
            contentType: false   // 告诉jQuery不要去设置Content-Type请求头
        })
        .success(function(data) {
            //代码
        });
    });
4、也可以用serializeArray函数模拟上面的formData对象，代码如下
    var allDatas = $("form").serializeArray();
    allDatas.push({name:'data',value: JSON.stringify(你的数据对象)});//追加的格式必须是name，value形式，打印allDatas的格式就知道了！！！
    $.post(url,allDatas,function(json){//代码
    });