非常基本,但作为一个新手,我正在努力。回声不显示任何值,只显示文本。我做错了什么?
Connect.php:
<?php
$connection = mysqli_connect('test.com.mysql', 'test_com_systems', 'systems');
if (!$connection){
die("Database Connection Failed" . mysqli_error($connection));
}
$select_db = mysqli_select_db($connection, 'swaut_com_systems');
if (!$select_db){
die("Database Selection Failed" . mysqli_error($connection));
}
?>
Get.php:
<?php
require('connect.php');
$query2 = "SELECT systemid FROM user WHERE username=test";
$result2 = mysqli_query($connection, $query2);
echo ( 'SystemID: '.$result2);
?>
答案 0 :(得分:3)
假设您已成功连接到数据库,则查询不正确。您必须将所有文本值包装在这样的引号中
<?php
require('connect.php');
$query2 = "SELECT systemid FROM user WHERE username='test'";
$result2 = mysqli_query($connection, $query2);
现在mysqli_query
将查询提交到运行它的数据库并构建结果集。要查看结果集,您需要使用fetch
函数之一从数据库中读取结果集,例如
$row = mysqli_fetch_assoc($result2);
echo 'SystemID: ' . $row['systemid'];
如果结果集中有多行,则必须在这样的循环中执行此操作
while ($row = mysqli_fetch_assoc($result2)){
echo 'SystemID: ' . $row['systemid'];
}
答案 1 :(得分:0)
您正在打印mysqli
结果object
。为了print
您必须使用的结果:
$row = mysqli_fetch_assoc($result2);
print_r($row);
答案 2 :(得分:0)
您需要使用以下内容收集mysqli_query
的结果:
require('connect.php');
$query2 = "SELECT systemid FROM user WHERE username=test";
$result2 = mysqli_query($connection, $query2);
while ($row = mysqli_fetch_assoc($result2))
{
echo "System ID is: " . $row['systemid'];
}