Question

我正在尝试通过其标题标题对表进行排序。但是我得到了无法解析的查询。这是我在MySQL中做错了吗？

我必须订购一个包含数千个记录的工作表。所以我只是在家里练习才能把它弄好。但是目前我似乎陷入了那个MySQL位。我按照不同网站上的提示进行了操作，这些网站上有这段代码，但仍然无法解释为什么这不起作用会有所帮助。

<html xmlns="http://www.w3.org/1999/xhtml">

<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <link href="../demo.css" rel="stylesheet" type="text/css" />
</head>
<body>

    <?php
    function makeHeaderLink($value, $key, $col, $dir) {
        $out = "<a href=\"" . $_SERVER['SCRIPT_NAME'] . "?c=";
        //set column query string value
        switch($key) {
            case "Acronym":
                $out .= "1";
                break;
            case "Full Name":
                $out .= "2";
                break;
            case "What":
                $out .= "3";
                break;
            default:
                $out .= "0";
        }

        $out .= "&d=";

        //reverse sort if the current column is clicked
        if($key == $col) {
            switch($dir) {
                case "ASC":
                    $out .= "1";
                    break;
                default:
                    $out .= "0";
            }
        }
        else {
            //pass on current sort direction
            switch($dir) {
                case "ASC":
                    $out .= "0";
                    break;
                default:
                    $out .= "1";
            }
        }

        //complete link
        $out .= "\">$value</a>";

        return $out;
    }

    switch($_GET['c']) {
        case "1":
            $col = "Acronym";
            break;
        case "2":
            $col = "Full_Name";
            break;
        case "3":
            $col = "What";
            break;
                }

    if($_GET['d'] == "1") {
        $dir = "DESC";
    }
    else {
        $dir = "ASC";
    }

    if(!$link = mysql_connect("localhost", "username", "password")) {
        echo "Cannot connect to db server";
    }
    elseif(!mysql_select_db("nazir_jargon")) {
        echo "Cannot select database";
    }
    else {

        if(!$rs = mysql_query("SELECT 'Acronym', 'Full Name', 'What' * FROM jargon1 ORDER BY $col $dir")) {
            echo "Cannot parse query";
        }
        elseif(mysql_num_rows($rs) == 0) {
            echo "No records found";
        }
        else {
            echo "<table class=\"bordered\" cellspacing=\"0\">\n";
            echo "<tr>";
            echo "<th>" . makeHeaderLink("Acronym", "Acronym", $col, $dir) . "</th>";
            echo "<th>" . makeHeaderLink("Full Name", "Full_Name", $col, $dir) . "</th>";
            echo "<th>" . makeHeaderLink("What", "What", $col, $dir) . "</th>";
            echo "</tr>\n";
            while($row = mysql_fetch_array($rs)) {
                echo "<tr><td>$row[person_id]</td><td>$row[Acronym]</td><td>$row[Full_Name]</td><td>$row[What]</td></tr>\n";
            }
            echo "</table><br />\n";
        }
    }
    ?>
</body>

Answer 1

删除单引号＆amp;来自查询的 * ，以便查询应如下所示。我希望它能奏效。

SELECT Acronym, Full Name, What FROM jargon1 ORDER BY $col $dir

如果不起作用，

可以打印

echo "SELECT Acronym, Full Name, What FROM jargon1 ORDER BY $col $dir";

让我知道你得到了什么？

注意：我认为你想使用反引号（`），但错误地使用单引号。

编辑1

如你所见

echo "SELECT Acronym, Full Name, What FROM jargon1 ORDER BY $col $dir";

打印

SELECT Acronym, Full Name, What FROM jargon1 ORDER BY ASC`

这意味着$col没有打印..请检查$col

出了什么问题

SELECT Acronym, Full Name, What FROM jargon1 ORDER BY columnNameMissing ASC`
                                                      ^^^^^^^^^^^^^^^^^

Answer 2

如果要指定返回的列，请从选择查询中删除*。否则用*来取回一切。一个或另一个。

PHP MySQL命令表代码错误“无法解析查询”

2 个答案:

编辑1