我试图从一个MySQL表中选择数据，但我得到以下错误消息之一:
Mysql_fetch_array()期望参数1为给定的资源布尔值
这是我的代码:
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');
while($row = mysql_fetch_array($result)) {
echo $row['FirstName'];
}
当前回答
试试这段代码，它工作得很好
将post变量赋给变量
$username = $_POST['uname'];
$password = $_POST['pass'];
$result = mysql_query('SELECT * FROM userData WHERE UserName LIKE $username');
if(!empty($result)){
while($row = mysql_fetch_array($result)){
echo $row['FirstName'];
}
}
其他回答
在MySQL查询之前包含一个连接字符串变量。例如，下面代码中的$ connect:
$results = mysql_query($connt, "SELECT * FROM users");
因为$username是一个PHP变量，我们需要将它作为字符串传递给mysqli，所以在查询中，你以单引号开始，我们将使用双引号，单引号和句号来连接("'.$username.'")，如果你以双引号开始，你将反转引号('".$username."')。
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE "'.$username.'"');
while($row = mysql_fetch_array($result))
{
echo $row['FirstName'];
}
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '".$username."' ");
while($row = mysql_fetch_array($result))
{
echo $row['FirstName'];
}
但是Mysql的使用已经贬值了很多，改用PDO。它很简单，但非常安全
不要使用已废弃的mysql_*函数(在php 5.5中已废弃的将在php 7中被删除)，你可以使用mysqli或pdo来实现
下面是完整的选择查询
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
// code here
}
} else {
echo "0 results";
}
$conn->close();
?>
试试下面的代码。它可能会工作得很好。
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName ='$username'");
while($row = mysql_fetch_array($result))
{
echo $row['FirstName'];
}
试试这段代码，它工作得很好
将post变量赋给变量
$username = $_POST['uname'];
$password = $_POST['pass'];
$result = mysql_query('SELECT * FROM userData WHERE UserName LIKE $username');
if(!empty($result)){
while($row = mysql_fetch_array($result)){
echo $row['FirstName'];
}
}
