【发布时间】:2014-10-10 06:47:40
【问题描述】:
这些是我的问题:
我从两个下拉列表中选择的值将变回默认值 当浏览器在表单中实现onchange函数时。
我要创建一个表,数据应该和 我从第二个下拉列表中选择的表格。
这是我的代码
PHP
<?php
$connectDatabase = mysql_connect("localhost","root","") or die(mysql_error());
$tables = array();
if(isset($_POST['select_db']))
{ // if its submitted
$select_db = $_POST['select_db'];
$mysql_select_db = mysql_select_db($select_db,$connectDatabase);
$drop_table = mysql_query("DROP TABLE pdf_table",$connectDatabase);
$query = "SHOW TABLES FROM $select_db";
$mysql_query = mysql_query($query,$connectDatabase);
while($row =mysql_fetch_assoc($mysql_query))
{
$tables[] = $row['Tables_in_' . $select_db]; // use associative instead
}
}
if(isset($_POST['select_table']))
{
$select_table = $_POST['select_table'];
$db = mysql_select_db($select_db,$connectDatabase);
$query_select = "Create Table pdf_table AS ( SELECT * FROM $select_table)";
$select_query = mysql_query($query_select,$connectDatabase);
}
?>
HTML 代码
<form class="Search_Form" action="moduleindex.php" method="POST">
<select name="select_db" onchange="this.form.submit();">
<option disabled selected>Select Database</option>
<option>section_masterfile</option>
</select>
<select onchange="this.form.submit();" name="select_table">
<option disabled selected>Select Table</option>
<?php foreach($tables as $table): ?>
<option value="<?php echo $table; ?>"><?php echo $table; ?></option>
<?php endforeach; ?>
</select>
</form>
【问题讨论】:
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创建表 mytable 作为选择...
标签: php html mysql select create-table