php - json_encode 不显示 HTML

Question

我正在从数据库中取一些数据，其中一些有HTML标签，如“<p>Hello</ p>”。但是，当逐步返回数据时，它会在“<p>Hello</p>”中显示 json_encode。我应该怎么做才能返回保存在数据库中的相同数据？

PHP:

$retorna = array(
    'conteudo_programatico' => $conteudo_programatico,
    'investimento' => $investimento,
    'coordenacao' => $coordenacao,
    'importante' => $importante
);

echo json_encode($retorna);

Ajax :

$(function(){
    $("button").click(function(){
        var id = $(this).attr("id");
        $.ajax({  
            type: "POST",
            data: {id: id},
            url: "paginas/ajax/form.php",
            dataType: "json",
            success: function(data){
                $("#cursos").slideDown();
                $(".call-conteudo").text(data['conteudo_programatico']);
                $(".call-investimento").text(data['investimento']);
                $(".call-coordenacao").text(data['coordenacao']);
                $(".call-importante").text(data['importante']);
            },
        });
    });
});

Answer 1

问题是您使用了 jQuery 的 .text() :

$(".call-conteudo").text(data['conteudo_programatico']);
// etc.

根据manual :

We need to be aware that this method escapes the string provided as necessary so that it will render correctly in HTML. To do so, it calls the DOM method .createTextNode(), does not interpret the string as HTML

...

The code $( "div.demo-container" ).text( "<p>This is a test.</p>" ); will produce the following DOM output:

 1 <div class="demo-container">
 2   &lt;p&gt;This is a test.&lt;/p&gt;
 3 </div>

你需要 html() :

$(".call-conteudo").html(data['conteudo_programatico']);
// etc.