php - 用 PHP 编写数据库的属性

Question

我正在编写一个用户可以在其中输入数据库名称的应用程序，我应该使用 PHP 将其所有内容写入表中。当我知道数据库名称时，我可以使用以下代码执行此操作。

$result = mysqli_query($con,"SELECT * FROM course");

echo "<table border='1'>
<tr>
<th>blablabla</th>
<th>blabla</th>
<th>blablabla</th>
<th>bla</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['blablabla'] . "</td>";
  echo "<td>" . $row['blabla'] . "</td>";
  echo "<td>" . $row['blablabla'] . "</td>";
  echo "<td>" . $row['bla'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

在这个例子中，我可以显示它，因为我知道表的名称是 course 并且它有 4 个属性。但是我希望能够显示结果，而不考虑用户输入的名称。所以如果用户想要查看讲师的内容应该有两列而不是 4 列。我怎样才能做到这一点。我用 html 获取表名。

Table:<input type="text" name="table">

编辑:Denis 的回答和 GrumpyCroutons 的回答都是正确的。你也可以问我，如果你不明白他们的解决方案中的某些内容。

Answer 1

快速写下它，对其进行评论(这样你就可以很容易地了解发生了什么，你看)，并为你测试它。

    <form method="GET">
        <input type="text" name="table">
    </form>

    <?php


    //can be done elsewhere, I used this for testing. vv
    $config = array(
         'SQL-Host' => '',
         'SQL-User' => '',
         'SQL-Pass' => '',
         'SQL-Database' => ''
    );
    $con = mysqli_connect($config['SQL-Host'], $config['SQL-User'], $config['SQL-Pass'], $config['SQL-Database']) or die("Error " . mysqli_error($con));
    //can be done elsewhere, I used this for testing. ^^


    if(!isSet($_GET['table'])) { //check if table choser form was submitted.
        //In my case, do nothing, but you could display a message saying something like no db chosen etc.
    } else {

    $table   = mysqli_real_escape_string($con, $_GET['table']); //escape it because it's an input, helps prevent sqlinjection.

    $sql     = "SELECT * FROM " . $table; // SELECT * returns a list of ALL column data
    $sql2    = "SHOW COLUMNS FROM " . $table; // SHOW COLUMNS FROM returns a list of columns

    $result  = mysqli_query($con, $sql);
    $Headers = mysqli_query($con, $sql2);

    //you could do more checks here to see if anything was returned, and display an error if not or whatever.

    echo "<table border='1'>";
    echo "<tr>"; //all in one row

    $headersList = array(); //create an empty array

    while($row = mysqli_fetch_array($Headers)) { //loop through table columns
         echo "<td>" . $row['Field'] . "</td>"; // list columns in TD's or TH's.
         array_push($headersList, $row['Field']); //Fill array with fields
    } //$row = mysqli_fetch_array($Headers)

    echo "</tr>";

    $amt = count($headersList); // How many headers are there?

    while($row = mysqli_fetch_array($result)) {
         echo "<tr>"; //each row gets its own tr
         for($x = 1; $x <= $amt; $x++) { //nested for loop, based on the $amt variable above, so you don't leave any columns out - should have been <= and not <, my bad
             echo "<td>" . $row[$headersList[$x]] . "</td>"; //Fill td's or th's with column data
        } //$x = 1; $x < $amt; $x++
             echo "</tr>";
    } //$row = mysqli_fetch_array($result)

    echo "</table>";
    }
    ?>