如何从MYSQL获取表名到组合框并将所选值分配给变量？ [英] How to get table's name from MYSQL to combobox and assign selected value to variable?

问题描述

您好，

我在PHP中遇到了使用combobox的问题。

我尝试将表的名称改为组合框，这是有效的。但我无法获得选定的值并分配给变量。请帮我。谢谢。

这是我的代码：

Hello,
I have a problem with combobox in PHP.
I try to get table's name to combobox, it's work. But I can't get selected value and assign to variable. Please help me. Thank you.
Here is my code:

$db= mysql_connect("localhost","root","");
	if(!$db)
	{
		echo "error";
		exit;
	}
$table = mysql_query("show tables from sctv_data2 LIKE '%nhap_%'");
echo "<select name = 'venue' >";
while (($row = mysql_fetch_row($table)) != null)
{
    echo "<option value = '{$row['0']}'";
    if ($selected_venue_id == $row['0'])
        echo "selected = 'selected'";
    echo ">{$row['0']}</option>";
}
echo "</select>";
if(isset($_POST['button']))
{
	$a =  $_POST['venue'];
	echo $a; // it's not print $a

推荐答案

db = mysql_connect（ localhost， root， ）;
if（！

db= mysql_connect("localhost","root",""); if(!

db）
{
echo 错误;
退出;
}

db) { echo "error"; exit; }

table = mysql_query（ 显示来自sctv_data2的表LIKE'％nhap _％'）;
echo < select name ='场地'>;
while （（

table = mysql_query("show tables from sctv_data2 LIKE '%nhap_%'"); echo "<select name = 'venue' >"; while ((

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