我做了一个php脚本,如下所示,运行它,输出是Record Added to Table
<?PHP
$user_name = "root";
$password = "";
$database = "test_db";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "INSERT INTO name (FirstName) VALUES ('bill')";
$result = mysql_query($SQL);
mysql_close($db_handle);
print "Records added to the database";
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>
现在查看表的内容,我做了
<?PHP
$user_name = "root";
$password = "";
$database = "test_db";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "Select * from name";
$result = mysql_query($SQL);
print "$result";
mysql_close($db_handle);
print "Records added to the database";
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>
但是输出是资源id #3Records添加到数据库,而我想看到表的内容
mysql_query的返回值不能像你尝试的那样发送到输出。
mysql_query返回一个结果集,例如用于http://www.php.net/mysql_fetch_row。
的例子:
while($row=mysql_fetch_array($result)) {
print_r($row);
}
$result
包含mysql_query
的返回值,该返回值为resource
(http://php.net/manual/en/function.mysql-query.php)
使用mysql_fetch_assoc($result)
或mysql_fetch_array($result)
或…获取实际数据
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
printf("Result: %s", $row[0]);
}
mysql_query()
将返回一个resultset对象,您必须在循环中读取该对象以获取其行:
while($row = mysql_fetch_array()) {
echo htmlspecialchars($row['column1']);
}
如果要从数据库中进行选择,请尝试:
<?PHP
$user_name = "root";
$password = "";
$database = "test_db";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "Select * from name";
$result = mysql_query($SQL);
while ($row = mysql_fetch_array($result))
{
print $row[0]." Record is selected";
}
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>