Question

您好我写下了以下PHP脚本，以HTML表格的形式显示在sqlite数据库中编写的数据。当我打开页面时，我只看到标题控制面板。我已将其保存为我的服务器根目录下的index.php。

<html>
    <head>
        <title>Switch and Status Report</title>
    </head>
    <body>
        <h1>Control Panel</h1>
        <table border="1" style="width:100%">
            <?php
                $dbhandle = sqlite_open('db/pj.db', 0666, $error);
                if (!$dbhandle) die ($error);
                $query = "SELECT * FROM state";
                $result = sqlite_query($dbhandle, $query);
                if (!$result) die("Cannot execute query.");
                $array = sqlite_fetch_all($result, SQLITE_NUM);
                echo($array);
                for ($x = 0; $x <= 6; $x++) {
                        $curarr = $array[$x];
                        $curpin = $curarr[0];
                        $curname = $curarr[1];
                        $curstate = $curarr[2];
                        $curdep = $curarr[3];
                        $textst = ""; 
                        $linkst = "";
                        $textdep = ""; 
                        $linkdep = "";
                        if($curstate==(0)){                             
                                $textst = "Turn On";
                                $linkst = "statechange.php";                        
                        }
                        elseif($curstate==(1)){
                                $textst = "Turn Off";
                                $linkst = "statechange.php";
                        }
                        if($curdep==(0)){                               
                                $textst = "Make Motion Dependent";
                                $linkst = "motionchange.php";                       
                        }
                        elseif($curstate==(1)){
                                $textst = "Make Motion Independent";
                                $linkst = "statechange.php";
                        }
                        echo("<tr>");
                        echo("<td>$curpin</td>");
                        echo("<td>$curname</td>");                      
                        echo("<td><a href='$linkst'>$textst</a></td>");             
                        echo("<td><a href='$linkdep'>$textdep</a></td>");                       
                        echo("</tr>");  
                        sqlite_close($dbhandle);                    
                } 
            ?> 
        </table>
    </body>
</html>

请排查并帮助我。提前谢谢。

Answer 1

$ curarr部分缺少你。在上面的第18行，看到你已经省略了$符号。它产生了一个错误。

<html>
<head>
    <title>Switch and Status Report</title>
</head>
<body>
    <h1>Control Panel</h1>
    <table border="1" style="width:100%">
        <?php
            $dbhandle = sqlite_open('db/pj.db', 0666, $error);
            if (!$dbhandle) die ($error);
            $query = "SELECT * FROM state";
            $result = sqlite_query($dbhandle, $query);
            if (!$result) die("Cannot execute query.");
            $array = sqlite_fetch_all($result, SQLITE_NUM);
            echo($array);
            for ($x = 0; $x <= 6; $x++) {
                    $curarr = $array[$x];
                    $curpin = $curarr[0];
                    $curname = $curarr[1];
                    $curstate = $curarr[2];
                    $curdep = $curarr[3];
                    $textst = ""; 
                    $linkst = "";
                    $textdep = ""; 
                    $linkdep = "";
                    if($curstate==(0)){                             
                            $textst = "Turn On";
                            $linkst = "statechange.php";                        
                    }
                    elseif($curstate==(1)){
                            $textst = "Turn Off";
                            $linkst = "statechange.php";
                    }
                    if($curdep==(0)){                               
                            $textst = "Make Motion Dependent";
                            $linkst = "motionchange.php";                       
                    }
                    elseif($curstate==(1)){
                            $textst = "Make Motion Independent";
                            $linkst = "statechange.php";
                    }
                    echo("<tr>");
                    echo("<td>$curpin</td>");
                    echo("<td>$curname</td>");                      
                    echo("<td><a href='$linkst'>$textst</a></td>");             
                    echo("<td><a href='$linkdep'>$textdep</a></td>");                       
                    echo("</tr>");  
                    sqlite_close($dbhandle);                    
            } 
        ?> 
    </table>
</body>

如何使用PHP将sqlite数据库中的数据制成表格

1 个答案: