Home/ Questions/Q 5978135
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-22T21:28:14+00:00

I’m designing a fairly simple reporting system. Right now using php(and later some Jquery)

  • 0

I’m designing a fairly simple reporting system. Right now using php(and later some Jquery) to let user log in and calculate totals and post to a database. My problems began of course when I tested the page in IE8. It has major problems with the echo <<<_END statement at line 75. Anyone know an alternate to coding the nested HTML in the PHP besides this. any help is appreciated.

<!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>
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    The Heredoc identifier must on a line of itself and must be the only thing on this line (including indentation and such). Remove the leading whitespace and make sure the newline is direct after _END;.

    However, in your case I suggest you to just leave the PHP-mode to output the plain html

    For example instead of

    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;

    this

    ?>
    <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

    Its usually more readable and many IDEs are capable to recognize it and highlight the “other stuff” as html.

    • 0