使用 PHP 从 MySQL DB 更新 HTML 组合框
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Updating an HTML combo box from MySQL DB using PHP
提问by vjrngn
need some help here. It's a very simple web app i'm developing but just needed some help with something.
这里需要一些帮助。这是我正在开发的一个非常简单的网络应用程序，但只是需要一些帮助。
Here's what's setup. I have a html form with one combo box. All I need is to update this combo box with the entries from a mysql table named 'supplier'. The input to this table 'supplier' is via another form on my website which i've already setup. I need help in auto updating this combo box from the table 'supplier'. Please let me know the php code for it. I've included my code as well. Thanks in advance! I have included the html form as well.
这是设置的内容。我有一个带有一个组合框的 html 表单。我所需要的只是使用名为“供应商”的 mysql 表中的条目更新此组合框。这个表“供应商”的输入是通过我网站上的另一个表格，我已经设置了。我需要帮助从表“供应商”自动更新此组合框。请让我知道它的php代码。我也包含了我的代码。提前致谢！我也包含了 html 表单。
采纳答案by Hitesh Siddhapura
replace your code
替换你的代码
$result = mysql_query("SELECT supplier FROM supplier");
while($row = mysql_fetch_array($result))
{
/*echo '<form action="">';*/
echo "<select name='supplier'>";
echo "<option value = '$row[supplier]'>""</option>";
echo "</select>";
with
和
$result = mysql_query("SELECT supplier FROM supplier");
echo "<select name='supplier'>";
while($row = mysql_fetch_assoc($result))
{
echo "<option value = '".$row[supplier]."'>".$row[supplier]."</option>";
}
echo "</select>";
回答by Micha El
So,
所以，
what is the problem? Is your script not working or do you want a active Combobox on one screen to be updated, when on another screen a new entry for supplier is entered?
问题是什么？当在另一个屏幕上输入供应商的新条目时，您的脚本是否不起作用，或者您是否希望更新一个屏幕上的活动组合框？
Ok, some hints:
好的，一些提示：
- The "select"-Tag should be placed outside the loop.
- Put the column name in "
- Add a display-Value for the options
- “select”-Tag 应该放在循环之外。
- 将列名放在“
- 为选项添加显示值
So, without checking it:
所以，不检查它：
/*echo '<form action="">';*/
echo "<select name='supplier'>";
while($row = mysql_fetch_array($result))
{
echo "<option value = '".$row["supplier"]."'>".$row["supplier"]."</option>";
}
echo "</select>";
回答by fantasitcalbeastly
Your problem is here:
你的问题在这里：
$result = mysql_query("SELECT supplier FROM supplier");
while($row = mysql_fetch_array($result))
{
/*echo '<form action="">';*/
echo "<select name='supplier'>";
echo "<option value = '$row[supplier]'>""</option>";
echo "</select>";
You're creating the drop down box (the Select) inside of the mysql data loop. As @Hitesh has explained. You need to create this outside of the loop and only echo out the data results within. For example:
您正在Selectmysql 数据循环内创建下拉框 (the )。正如@Hitesh 所解释的那样。您需要在循环之外创建它，并且只回显其中的数据结果。例如：
$result = mysql_query("SELECT supplier FROM supplier");
echo "<select name='supplier'>";
while($row = mysql_fetch_array($result))
{
echo "<option value=".$row['supplier'].">".$row['supplier']."</option>";
}
echo "</select>";
This will output your drop down box, with all your supplier names as the value and the displayed text options.
这将输出您的下拉框，其中包含所有供应商名称作为值和显示的文本选项。
If you attempted to do the following:
如果您尝试执行以下操作：
$result = mysql_query("SELECT supplier FROM supplier");
while($row = mysql_fetch_array($result))
{
/*echo '<form action="">';*/
echo "<select name='supplier'>";
echo "<option value = '$row[supplier]'>""</option>";
echo "</select>";
}
You would just end up with as many drop down boxes as you had suppliers in your database. This is because you're creating a new Selectbox for each record found.
您最终会得到与数据库中的供应商一样多的下拉框。这是因为您要Select为找到的每条记录创建一个新框。
Also, without specifying the second $row['supplier']between the optiontags, you'd just end up with a blank (empty) drop down box.
此外，如果不指定标签$row['supplier']之间的第二个option，您只会得到一个空白（空）下拉框。
Hope this helps.
希望这可以帮助。
回答by Neeraj Singh
Here is complete code:
这是完整的代码：
PHP Code:
PHP代码：
<?php
// select box open tag
$selectBoxOpen = "<select name='supplier'>";
// select box close tag
$selectBoxClose = "</select>";
// select box option tag
$selectBoxOption = '';
// connect mysql server
$con = mysql_connect("localhost","user","pass");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
// select database
mysql_select_db("rtgs", $con);
// fire mysql query
$result = mysql_query("SELECT supplier FROM supplier");
// play with return result array
while($row = mysql_fetch_array($result)){
$selectBoxOption .="<option value = '".$row['supplier']."'>".$row['supplier']."</option>";
}
// create select box tag with mysql result
$selectBox = $selectBoxOpen.$selectBoxOption.$selectBoxClose;
?>
HTML Code:
HTML代码：
<form action="ntxn.php" method="post">
<table>
<tr>
<td>Supplier Name:</td>
<td>
<?php echo $selectBox;?>
</td>
</tr>
<tr>
<td>Bill No. :</td>
<td><input type ="text" name="billno"/></td>
</tr>
<tr>
<td>Bill Date : </td>
<td><input type="date" name="billdate"/></td>
</tr>
<tr>
<td>Bill Amount : </td>
<td><input type="text" name="billamt"/></td>
</tr>
<tr>
<td>NEFT / RTGS No. :</td>
<td><input type="text" name="rtgs"/></td>
</tr>
<tr>
<td><input type="submit" name="Save"/></td>
</tr>
</table>
</form>
