我创建了要求用户以添加一个特定的行: 名字,姓氏,地点,状态回声从MySQL数据库
一段时间后,我已经收到5个输入。 mysql表名为users和表格数据如下:
id first_name location status
== ========== ======== ========
1 Chris UK Married
2 Anton Spain Single
3 Jacob UK Single
4 Mike Greece Married
5 George UK Married
此外,我通过POST方法接收输入的方式。所以:
$firstname=$_POST['FIRST_NAME']; // First Name: <input type="text" name="FIRST_NAME">
$location=$_POST['LOCATION']; // Location: <input type="text" name="LOCATION">
$status=$_POST['STATUS']; // Status: <input type="text" name="STATUS">
我创建一个查询,选择从英国所有用户都结婚了:
$query = "SELECT * FROM users WHERE location='UK' AND status='Married'";
$result = mysqli_query($dbc,$query); //$dbc is the connection to my database
$row = mysqli_fetch_array($results, MYSQLI_BOTH);
换句话说:
id first_name location status
== ========== ======== ========
1 Chris UK Married
5 George UK Married
问题:
1) $ row数组如下所示:
$row= array(
array(1 Chris UK Married),
array(5 George UK Married)
);
2)如何在实现筛选后回显数据库的内容WHERE location ='UK'AND status ='Married'?
我需要它是这样的:
Hello Chris! You are from UK and you are married!
Hello George! You are from UK and you are married!
我知道,我必须使用foreach循环(回声阵列),但我已经尝试过了,它不的我试过的东西work.One是这是我在php.net发现:
与列表()开箱嵌套数组
(PHP 5> = 5.5.0)
PHP 5.5添加到迭代能力一个数组数组,并通过提供一个list()作为值将嵌套数组解压缩成循环变量。
例如:
<?php
$array = [
[1, 2],
[3, 4],
];
foreach ($array as list($a, $b)) {
// $a contains the first element of the nested array,
// and $b contains the second element.
echo "A: $a; B: $b\n";
}
?>
当我使用上述我收到以下错误:
Parse error: syntax error, unexpected T_LIST in C:\wamp\www....
任何建议?
据我所知,我不知有该ID与其他数据预先链接..
感谢。
你需要循环用于获取所有数据,而($行= mysqli_fetch_array($结果,MYSQLI_BOTH)){回声$行[“身份证” ]。 //所有其他} –
对于问题1:如果你想知道某些东西的结构,你可以使用['var_dump'](http://php.net/var_dump),例如。 'var_dump($ row)' – TiiJ7