mysql_query()期望参数1为字符串,但字符串存在

时间:2013-08-26 18:08:57

标签: php mysql

我是mysql的新手,我在godaddy帐户中使用phpmyadmin创建了一个表,但我无法通过php添加任何内容。代码是:

$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';

mysql_query($conn,"INSERT INTO entries (business) VALUES ('Test biz')");

我成功连接但是我收到错误“mysql_query()期望参数1为字符串,资源给出”就行了

mysql_query($conn,"INSERT INTO entries (business) VALUES ('Test biz')");

关于为什么会这样的概念?谢谢!

3 个答案:

答案 0 :(得分:0)

需要改变

mysql_query($conn,"INSERT INTO entries (business) VALUES ('Test biz')");

mysql_query("INSERT INTO `entries` ( `business` ) VALUES ('Test biz')");

这是manual

答案 1 :(得分:0)

<强>解决方案

没有必要在你的mysql查询中传递$ conn,试试没有它,

有关语法的详细信息,请参阅此页面:http://php.net/manual/en/function.mysql-query.php

答案 2 :(得分:0)

你需要调用mysql_select_db($ conn,'');

$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';

mysql_query($ conn,“INSERT INTO entries(business)VALUES('Test biz')”);

$conn = mysql_connect($dbhost, $dbuser, $dbpass);

if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_select_db('<your db name here>',$conn);

mysql_query("INSERT INTO entries (business) VALUES ('Test biz')");

您不需要将$ conn传递给mysql_query();

由于