使用php中的选择显示我的数据库表

Question

我正在尝试使用此代码在数据库中显示所有表格。

<label>
  <select class="selectpicker" name="tabel">

        <?php
        include "koneksi.php";
        $koneksi ->select_db('testskripsi');
        $sql =  mysqli_query($koneksi, "show tables") ;
        while($table = $sql->fetch_array(MYSQLI_ASSOC)) {
            echo '<option name="opsi" value = "' . $table['Tables'] . '">' . $table['Tables'] . '</option>';
        }

        ?>
    </select>
</label>

这是我的结果

我想我不知道如何填充这个数组来显示我的表格

' . $table['Tables'] . '

Answer 1

请尝试使用Tables_in_testskripsi而不是Tables.here是您的正确代码。

<label>
<select class="selectpicker" name="tabel">

    <?php
    include "koneksi.php";
    $koneksi ->select_db('testskripsi');
    $sql =  mysqli_query($koneksi, "show tables") ;
    while($table = $sql->fetch_array(MYSQLI_ASSOC)) {
        echo '<option name="opsi" value = "' . $table['Tables_in_testskripsi'] . '">' . $table['Tables_in_testskripsi'] . '</option>';
    }

    ?>
</select>
</label>

让我知道它是否对你有帮助

Answer 2

获取表名，

$tables=$this->db->query("
  SELECT t.TABLE_NAME AS myTables
  FROM INFORMATION_SCHEMA.TABLES AS t
  WHERE t.TABLE_SCHEMA = 'database name'
  AND t.TABLE_NAME LIKE '%a%' 
")->result_array();

foreach($tables as $key => $val) {
  echo $val['myTables']."<br>";// myTables is the alias used in query.
}