[英]mysql_num_rows() Error
<html>
<form action="update.php" method="POST" name="ID">
<input type="text" name="ID"">
<input type="Submit" value="Submit">
</form>
</html>
Up there is the submit form to get an ID number. 上面有提交表单以获取ID号。 I try to get that ID entered by user ( NOTE: It's a number) and show mysql table row coresponding to that ID.
我尝试让用户输入该ID(注意:这是一个数字),并显示与该ID对应的mysql表行。 Example : User enter 2 and row number 2 from database is shown.
示例:用户输入2,显示数据库中的行号2。 My problem is that all rows are shown and not only wanted one.
我的问题是显示了所有行,而不仅仅是想要一个。 - Extra Question : How can I show user an error if he entered a NULL value ?
-额外问题:如果用户输入NULL值,如何显示错误?
<?php
$id=$_POST['ID'];
.
.
.
mysql_connect($host,$username,$password);
if (!mysql_select_db($database))
die("Can't select database");
$query="SELECT * FROM table WHERE ID= '$id'";
$result = mysql_query("SELECT * FROM vbots");
$num=mysql_num_rows($result) or die("Error: ". mysql_error(). " with query ". $query);
mysql_close();
.
.
.
?>
You're not running your query. 您没有运行查询。
You have this: 你有这个:
$query="SELECT FROM table WHERE ID= '$id'";
$result = mysql_query("SELECT * FROM vbots");
You want this: 你要这个:
$query="SELECT FROM table WHERE ID= '$id'";
$result = mysql_query( $query);
** Insert nag about SQL injection ** ** 插入有关SQL注入的内容 **
It should be this. 应该是这个
<?php
$id=$_POST['ID'];
mysql_connect($host,$username,$password);
if (!mysql_select_db($database))
die("Can't select database");
$query="SELECT * FROM table WHERE ID= '$id'";
$result = mysql_query($query);
$num=mysql_num_rows($result) or die("Error: ". mysql_error(). " with query ". $query);
mysql_close();
?>
检查$ id是否为空:if(!isset($ id))die(“输入ID值!”);
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