Question

当我选择从我的数据库中检索某些数据的选项时，我收到以下错误消息：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1.

为什么？第1行什么都没有？另外，你能教我如何搜索检索到的数据吗？

实施例。如果我选择选项1＆＃34; Antal narvaro＆＃34;然后按确定按钮，以下sql-query将执行：SELECT * FROM mytable WHERE fran = 'besok';

一旦我检索到这些数据，我想进一步在其中搜索某些信息，如姓名，年龄等。否则，我很感激我可以通过在下拉列表中选择选项然后按OK按钮来执行以下SQL查询。

我非常擅长 PHP 和HTML，提前感谢！

<html>
    <head> 
        <title>AiFind</title>
        <link rel="stylesheet" href="Style.css">
        <script type="text/javascript" src="logic.js"></script>


    </head>
    <body>
        <h1><img src="https://imagizer.imageshack.us/v2/278x186q90/844/k32d.png" 
                 height="70" width="100" alt="Aifind-logo copy"></h1>
    </body>

</html>


<?php

include "connection.php";

$sql = "SELECT";

if (isset($_POST['search'])) {

    $search_term = mysql_real_escape_string($_POST['search_box']);

    switch($_POST['filter']) {

        Default: 
            $sql .= " ";
            break;
        case 1:
            $sql .= " * from mytable WHERE fran = 'Besök'; ";
            break;
        case 2:
            $sql .= " * from mytable WHERE fran = 'Bok'; ";
            break;
        case 3:
            $sql .= " Postort, COUNT(*) FROM mytable WHERE AAA_diam != ";
            $sql .= " 'missing' GROUP BY Postort; ";
            break;
        case 4:
            $sql .= " *, COUNT(*) FROM mytable WHERE AAA_diam = 'missing'";
            $sql .= " GROUP BY Postort";
            break;
        case 5:
            $sql .= " COUNT(*) AS Antal_over_30 FROM mytable WHERE ";
            $sql .= " AAA_diam > 30 AND AAA_diam != 'missing' AND ";
            $sql .= " AAA_diam != 'n.a' AND AAA_diam != 'Missing'";
            break;
        case 6:
            $sql .= " COUNT(*) AS Antal_over_30 FROM mytable WHERE ";
            $sql .= " AAA_diam > 50 AND AAA_diam != 'missing' AND ";
            $sql .= " AAA_diam != 'n.a' AND AAA_diam != 'Missing'";
            break;
        case 7:
            $sql .=  "";
            break;

    }

}


$query = mysql_query($sql) or die(mysql_error());


?>


<!--Sökrutan där sökord matas in -->
<form name="Select_filter" method="POST" action="VGR_data_display.php">

    <select id="dropdown" name="filter">
        <option value=""></option>
        <option value="1">Antal narvaro</option>
        <option value="2">Antal franvaro</option>
        <option value="3">Antal narvaro: destination</option>
        <option value="4">Antal franvaro: destination</option>
        <option value="5">Antal A_diam > 3cm</option>
        <option value="6">Antal A_diam > 5cm</option>
        <option value="7">Antal franvaro: procent</option>

    </select>

    <!--search bar for search term input -->
    <input id="search_box" type="text" name="search_box" value="" />
    <input id="submit" type ="submit" name ="search" value ="Ok">

</form>

<table style="margin:auto;" id="table" border='1'>
<tr>
<th>ID</th>
<th>F_ar</th>
<th>Postnr</th>
<th>Postort</th>
<th>Vardgivare</th>
<th>Team</th>
<th>Orsak</th>
<th>Planerat_datum</th>
<th>fran</th>
<th>AAA_diam</th>
</tr>


<?php while($row = mysql_fetch_array($query)) { ?>
  <tr>
  <td><?php echo $row['id']; ?></td>
  <td><?php echo $row['F_ar']; ?></td>
  <td><?php echo $row['Postnr']; ?></td>
  <td><?php echo $row['Postort']; ?></td>
  <td><?php echo $row['Vardgivare']; ?></td>
  <td><?php echo $row['Team']; ?></td>
  <td><?php echo $row['Orsak']; ?></td>
  <td><?php echo $row['Planerat_datum']; ?></td>
  <td><?php echo $row['fran']; ?></td>
  <td><?php echo $row['AAA_diam']; ?></td>
  </tr>

<?php } ?>

尝试使用下拉列表中的选项从数据库中获取数据时，mysql语法错误

0 个答案: