如何输出id为31的行

Question

<?php
$link = mysql_connect('localhost', 'username', 'password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';

if (!mysql_select_db('database'))
    die("Can't select database");


// choose id 31 from table users

echo $id; //31
echo $name; //id31's name
echo $surname //id31's surname
echo $blablabla //id31's blablabla

mysql_close($link);
?>

Answer 1

试一试:

$id = mysql_real_escape_string(31);
$result = mysql_query('SELECT * FROM `users` where `id` = ' . $id . ' LIMIT 1');

$row = mysql_fetch_assoc($result);

echo $row['id']; // 31
echo $row['name']; // name
// etc..

// If you wanted to output ALL the fields you could loop through
foreach($row as $field_name => $value) {
  echo $field_name . ': ' . $value . '<br />';
}

// The output would look something like
// id: 31
// name: John Smith
// ...

使用的功能:

• mysql_real_escape_string() - 转义字符串中的特殊字符以用于SQL语句
• mysql_query() - 发送MySQL查询
• mysql_fetch_assoc() - 将结果行作为关联数组获取