php中的按钮在while循环中时如何给出规范 [英] how to give specification for button in php when it is in while loop
问题描述
<?php
echo "<html><form name=form1 method=post>";
$c = mysql_connect("localhost","root","");
mysql_select_db("test", $c);
$query = mysql_query("select * from age");
while ($s = mysql_fetch_row($query))
{
echo "<table><tr><td>Name :<textarea rows=5 cols=5 name=name>$s[0] </textarea></td></tr><tr><td>Age :<textarea rows=5 cols=5 name=age>$s[1] </textarea></td></tr><tr><td><input type=submit value='click' name=submit></td></tr></table>";
}
if(isset($_POST['submit']))
{
$name = $_POST['name'];
$age = $_POST['age'];
echo "name is $name"." age is $age";
}
echo "</form></html>";
?>
在测试表中，有 4 个条目，其中包含姓名和年龄....我正在获取该姓名和年龄并将其放入 textarea 中，然后我为每个表创建了一个按钮...同时单击该按钮我得到了数据库中的最后一个条目..但我想要相应的姓名和年龄..请帮帮我
here in the test table there are 4 entries present with name and age.... I am fetching that name and age and putting that in textarea and I created one button for each table.. while click that button I am getting last entry in database.. but I want corresponding name and age.. please help me
推荐答案
目前，您正在为所有条目使用一个大的 元素.发送该表单时，实际值会被多次覆盖，而您只会看到最后一个.
Currently you are using one big <form> element for all your entries. When sending that form, the actual values get overridden many times and you just see the last one.
为了解决这个问题，你可以，例如，用这样的每个条目制作一个表格:
To solve this, you could, e.g., make a form out of every entry like this:
echo "<html>";
$c = mysql_connect("localhost", "root", "");
mysql_select_db("test", $c);
$query = mysql_query("select * from age");
while ($s = mysql_fetch_row($query))
{
echo "<form method=post><table><tr><td>Name :<textarea rows=5 cols=5 name=name>$s[0] </textarea></td></tr><tr><td>Age :<textarea rows=5 cols=5 name=age>$s[1] </textarea></td></tr><tr><td><input type=submit value='click' name=submit></td></tr></table></form>";
}
if (isset($_POST['submit']))
{
$name = $_POST['name'];
$age=$_POST['age'];
echo "name is $name";
echo "age is $age";
}
// ... rest of the code ...
无论如何，您使用的 mysql 函数已被弃用.看看 PDO 或 mysqli 作为替代！
Anyway, the mysql functions, you use, are deprecated. Have a look at PDO or mysqli as a replacement!
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