Question

我正在设计一个相当简单的报告系统。现在使用php（以及后来的一些Jquery）让用户登录并计算总数并发布到数据库。当我在IE8中测试页面时，我的问题开始了。它在第75行的echo <<<_END语句中存在重大问题。除此之外，任何人都知道在PHP中编写嵌套HTML的替代方法。任何帮助表示赞赏。

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
   "http://www.w3.org/TR/html4/strict.dtd">
<HTML>
   <HEAD>
        <link rel="stylesheet" type="text/css" href="OPSstyle.css" />
 </HEAD>
   <BODY>
    <div id="wrapper">



        <div id="bodyContent">


        <div id="header">
        <img id="logo" src="OPS_logo.gif" alt="OPS Logo"/>
            <p>OPS ASSESSMENT</p>

        </div>

        <div id="leftNav">
            <p>To begin please login</p>
            <a href="loginusers.php">Test Login</a><br/>
            <a href="sqltest.php">Form Test</a><br/>

        </div>

        <div id="content">
        <?php

        //Connects to database
        require_once 'login.php';
        $db_server = mysql_connect($db_hostname, $db_username, $db_password);
            if(!$db_server) die("Unable to connect to MySQL: " .mysql_error());

        mysql_select_db($db_database)
            or die ("unable to select database: " .mysql_error());

        //deletes record
        if(isset($_POST['delete']) && isset($_POST['AssessmentID']))
        {
            $AssessmentID = get_post('AssessmentID');
            $query2 = "DELETE FROM assessmentscores WHERE AssessmentID='$AssessmentID'";


            if(!mysql_query($query2, $db_server))
                echo "Delete Failed: $query(br />" .
                mysql_error() . "<br /><br />";
        }

        //Inserts record
        if (isset ($_POST['AssessmentID'])&&
                    isset($_POST['Date']) &&
                    isset($_POST['Inspector']) &&
                    isset($_POST['PlantAssist']) &&
                    isset($_POST['Safety_Total']) 
                    )
                {
                    $AssessmentID = get_post('AssessmentID');
                    $Date = get_post('Date');
                    $Inspector = get_post('Inspector');
                    $PlantAssist =get_post('PlantAssist');
                    $Safety_Total = get_post('Safety_Total');
;

                    $query = "INSERT INTO opsassessment.assessmentscores(AssessmentID, Date, Inspector, PlantAssist, `Safety_Total`) VALUES" .
                        "('$AssessmentID', '$Date', '$Inspector', '$PlantAssist', '$Safety_Total')";

                    if (!mysql_query($query, $db_server))
                        echo "INSERT failed: $query<br />" .
                        mysql_error() . "<br /><br />";
                }

    echo <<<_END
        <form action = "sqltest.php" method="post"><pre>
        AssessmentID    <input type="text" name="AssessmentID" /><br>
        Date        <input type="text" name="Date" /><br>
        Inspector   <input type="text" name="Inspector" /><br>
        PlantAssist <input type="text" name="PlantAssist" /><br>
        Safety_Total    <input type="text" name="Safety_Total" /><br>

                <input type="submit" value="ADD RECORD" /><br>
        </pre></form>
    _END;

        $query = "SELECT * FROM assessmentscores";
        $result = mysql_query($query);

        if (!$result) die("Database access failed: " .mysql_error());

        $rows = mysql_num_rows($result);

        for($j = 0; $j < $rows; ++$j)
        {
            $row= mysql_fetch_row($result);
    echo <<<_END
            <pre>
            AssessmentID:   $row[0]
            Date:           $row[1]
            Inspector:      $row[2] 
            PlantAssist:        $row[3]
            Safety_Total:       $row[4]

            </pre>

            <form action="sqltest.php" method="post">
            <input type ="hidden" name="delete" value ="yes" />
            <input type = "hidden" name="AssessmentID" value = "$row[0]" />
            <input type ="submit" value="DELETE RECORD" /></form>
    _END;
        }

        mysql_close($db_server);

        function get_post($var)
        {
            return mysql_real_escape_string($_POST[$var]);
        }


        ?>

        </div>


        <a href="form1.php">Login Form</a><br><br>

        <a href="testOPS.php">OPS Assessment</a>

        </div>
</body>
</html>

Answer 1

Heredoc标识符必须在其自身的一行上，并且必须是此行上的唯一内容（包括缩进等）。删除前导空格并确保换行符在_END;后直接显示。

但是，在你的情况下，我建议你离开PHP模式输出普通的html

例如，而不是

echo <<<_END
    <form action = "sqltest.php" method="post"><pre>
    AssessmentID    <input type="text" name="AssessmentID" /><br>
    Date        <input type="text" name="Date" /><br>
    Inspector   <input type="text" name="Inspector" /><br>
    PlantAssist <input type="text" name="PlantAssist" /><br>
    Safety_Total    <input type="text" name="Safety_Total" /><br>

            <input type="submit" value="ADD RECORD" /><br>
    </pre></form>
_END;

此

?>
    <form action = "sqltest.php" method="post"><pre>
    AssessmentID    <input type="text" name="AssessmentID" /><br>
    Date        <input type="text" name="Date" /><br>
    Inspector   <input type="text" name="Inspector" /><br>
    PlantAssist <input type="text" name="PlantAssist" /><br>
    Safety_Total    <input type="text" name="Safety_Total" /><br>

            <input type="submit" value="ADD RECORD" /><br>
    </pre></form>
<?php

它通常更具可读性，许多IDE能够识别它并突出显示“其他内容”为html。

Answer 2

正如评论者所说，HEREDOC（＆lt;＆lt;＆lt;和_END;）位与服务器端PHP相关，浏览器应该对它们一无所知。

HEREDOC标签的一个主要的，易于忘记的黄金法则是，结束标记不能被隐藏，永远！或者它将会中断。

下面是关于他们的读物：

http://www.php.net/manual/en/language.types.string.php#language.types.string.syntax.heredoc

如果它真的破坏了它，它真的就是这样做（虽然，它没有理由应该这样） - 记住它只是一个输出块，所以你可以用它来代替它：

$out = "blah blah";
$out .= "more blah blah";
$out .= "more blah blah";
$out .= "more blah blah";
$out .= "more blah blah";

echo $out;

IE8中＆lt;＆lt;＆lt; _END php标记的问题

2 个答案: