Question

这里我在php中嵌入了html代码。

但是执行时会给出错误。我查了很多次但看不到bug。帮助赞赏：请看

   <?php
    $url = $_POST['url'];
    $user_id = $_POST ['userid'];
    if(isset($_POST['rate']))
    {
        $rate =$_POST['rate'];
    }
    else
        $rate = 0;
    $data = file_get_contents($url);
    function get_title($html) 
    {
        return preg_match('!<title>(.*?)</title>!i', $html, $matches) ? $matches[1] : '';
    }

    function get_logo($html) 
    {
        preg_match_all('/\bhttps?:\/\/\S+(?:png|jpg)\b/', $html, $matches);
        //echo "mactch : $matches[0][0]";
        return $matches[0][0];  
    }   

    function plaintext($html)
    {
        // remove comments and any content found in the the comment area (strip_tags only removes the actual tags).
        $plaintext = preg_replace('#<!--.*?-->#s', '', $html);

        // put a space between list items (strip_tags just removes the tags).
            $plaintext = preg_replace('#</li>#', ' </li>', $plaintext);

            // remove all script and style tags
        $plaintext = preg_replace('#<(script|style)\b[^>]*>(.*?)</(script|style)>#is', "", $plaintext);

        // remove br tags (missed by strip_tags)
            $plaintext = preg_replace("#<br[^>]*?>#", " ", $plaintext);

            // remove all remaining html
            $plaintext = strip_tags($plaintext);

        return $plaintext;
    }

    function trim_display($size,$string)
    {
        $trim_string = substr($string, 0, $size);

        $trim_string = $trim_string . "...";
        return $trim_string;
    }

    $title = get_title($data);
    $logo = get_logo($data);
    $title_display = trim_display(30,$title);
    $content = plaintext($data); 
    $Preview = trim_display(100,$content); //to Show first 100 char of the web page as preview

    function MakeTinyUrl($url)
    {
        return md5($url);
    }

    $hash = MakeTinyUrl($url);
    ob_start(); 
    $con = mysqli_connect('127.0.0.1', 'root', '', 'mysql');
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
        return;
    }

    $content=mysqli_real_escape_string($con,$content);
    $Preview=mysqli_real_escape_string($con,$Preview);
    $title_display=mysqli_real_escape_string($con,$title_display);

    $result = mysqli_query($con,"SELECT COUNT(*) as val FROM post_data WHERE url ='".$url."' and userid='".$user_id."'");
    $bool= mysqli_fetch_assoc($result); 
    if($bool['val'])
    {
        echo '<div style="clear:both;"><i>You have already worked on this url..</i> </div>';
    }
    else
    {
        $insertQuery = "INSERT INTO post_data(`userid`, `url`, `desc`, `preview`, `img_url`, `title` ,`hash`) VALUES ('".$user_id."','".$url."','".$content."','".$Preview."','".$logo."','".$title_display."','".$hash."')";
        if (!mysqli_query($con,$insertQuery))
        {
            die('Error: ' . mysqli_error($con));
        }
    }

    $result = mysqli_query($con,"SELECT * FROM post_data WHERE userid ='".$user_id."' and url='".$url."'");
    //This will fetch only one row from db
    while ($row = @mysqli_fetch_array($result))
    {
        $title = $row['title'];
        $url = $row['url'];
        $preview = $row['preview'];
        $image = $row['img_url'];       
    }
    //Update Rate value in table
    $update = "update post_data set rate='".$rate."' where url='".$url."'";
    if (mysqli_query($con, $update))
    {
        //echo "updated";
    }
    else
    {
        //echo "Not updated";
    } 



echo    '<style type="text/css">
    .fragment 
    {
        font-size: 12px;
        font-family: tahoma;
        height: 140px;
        width: 400px;
        border: 1px solid #ccc;
        color: #555;
        display: block;
        padding: 10px;
        box-sizing: border-box;
        text-decoration: none;
    }

    .fragment:hover 
    {
        box-shadow: 2px 2px 5px rgba(0,0,0,.2);
    }

    .fragment img 
    {
        float: left;
        margin-right: 10px;
    }

    .fragment h3 
    {
        padding: 0;
        margin: 0;
        color: #369;
    }

    .fragment h4 
    {
        padding: 0;
        margin: 0;
        color: #000;
    }

    .fragment tbox 
    {
        padding: 1px;
        height: 30px;
        width: 424px;
    }
    #close 
    {
        float:right;
        display:inline-block;
        padding:2px 5px;
        background:#ccc;
    }

    </style>

    <a class="fragment" href="google.com">
        <div>
            <span id="close" onclick="this.parentNode.parentNode.parentNode.removeChild(this.parentNode.parentNode); return false;">x</span>';
            echo '<img src =' $image echo '" height="116" width="116" alt="some description"/>';
            echo '<h3>'; $title echo '</h3>';
            echo '<h4>'; $url echo ' </h4>';
            echo '<p class="text">';  $preview 
                echo '</p>

        </div>
    </a>';


        $div1 = ob_get_clean();
        ob_start(); 
        $result = mysqli_query($con,"SELECT * FROM post_data WHERE userid ='".$user_id."'");
        echo "Records for user : $user_id"; echo "<br/>";
        while ($row = @mysqli_fetch_array($result))
        {
            $title = $row['title'];
            $url = $row['url'];
            $preview = $row['preview'];
            $image = $row['img_url'];       
            echo "Title: $title"; echo '<br/>';
            echo "URL: $url";  echo '<br/>';
            echo "Preview : $preview"; echo '<br/>';
            echo "Image url: $image"; echo '<br/>';
            //echo '</br>';
        }     
        $div2 = ob_get_clean();
        $resultArray = array("resultOne" => $div1,"resultTwo" => $div2);
        echo json_encode($resultArray);

    ?>

https://gist.github.com/karimkhanp/b8bfb88db6e438fb8be6

当我尝试获取json响应时，它会给出以下错误：

SyntaxError: JSON.parse: unexpected character


var data=JSON.parse(xmlhttp.responseText);

Answer 1

使用字符串连接（.）来连接字符串。

只需替换168-176行：

<a class="fragment" href="google.com">
        <div>
            <span id="close" onclick="this.parentNode.parentNode.parentNode.removeChild(this.parentNode.parentNode); return false;">x</span>';
            echo '<img src ='.$image.'" height="116" width="116" alt="some description"/>';
            echo '<h3>'.$title.'</h3>';
            echo '<h4>'.$url.'</h4>';
            echo '<p class="text">'.$preview.'</p>
        </div>
    </a>';

我建议您使用更好的文本编辑器，如Notepad ++或Adobe Dreamweaver。

在php中嵌入html代码

1 个答案: