我要选择一个表,并创建一个表,从表中的相同数据,我选择
问题描述:
这是我的问题:我要选择一个表,并创建一个表,从表中的相同数据,我选择
我从两个选择的值下拉列表中就会变回其默认值为 当浏览器在窗体中实现函数onchange时。
我想创建一个表格,数据应该与我从第二个下拉列表中选择的表格 相同。
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 $table; ?></option>
<?php endforeach; ?>
</select>
</form>
答
添加支票foreach循环而产生的选项,如果某个值发布和价值等于t Ø循环值
<select onchange="this.form.submit();" name="select_table">
<option disabled selected>Select Table</option>
<?php foreach($tables as $table){?>
<?php
$selected ="";
if(isset($_POST['select_table']) && $_POST['select_table'] == $table){
$selected = 'selected="selected"';
} ?>
<option value="<?php echo $table; ?>" <?php echo $selected ;?> ><?php echo $table; ?></option>
<?php }?>
</select>
创建表mytable的AS SELECT ... – Strawberry 2014-10-10 06:58:13