2015-04-07 106 views
0

我尝试制作一个简单的页面查看计数器。 每次页面刷新时,数字应增加+1。Php mysql页面查看计数器

当我在Phpmyadmin中执行代码时,它一切正常。但在PHP中,计数器返回,但不增加。我做错了什么?

<?php 
$username = "username"; 
$password = "password"; 
$hostname = "hostname"; 
$dbname = "dbname"; 


// Create connection 
$conn = new mysqli($hostname, $username, $password, $dbname); 
// Check connection 
if ($conn->connect_error) { 
die("Connection failed: " . $conn->connect_error); 
} 

$sql = "UPDATE `Games` SET `Played`= `Played`+1 WHERE 'ID'='3'"; 

$sql = "SELECT `Played` FROM `Games` WHERE `ID` = 3"; 

$result = mysqli_query($conn, $sql); 

if (mysqli_num_rows($result) > 0) { 
// output data of each row 
while($row = mysqli_fetch_assoc($result)) { 
    echo "" . $row[Played]. "<br>"; 
} 
} else { 
echo "0 results"; 
} 

$conn->close(); 
?> 
+2

这是一个错误? ''密码;'<=缺少引号,另外,这是一个'WHERE'ID''不是数值的列 –

+0

删除''ID''上的单引号 – Daan

+0

'$ result = mysqli_query($ conn,$ sql)或者死(mysqli_error($ conn));'会发出语法错误信号 –

回答

1

您必须创建更新串入$sql变量之后做出mysqli_query电话。你现在正在做的只是用SELECT查询字符串覆盖它。

[...] 

$sql = "UPDATE `Games` SET `Played`= `Played`+1 WHERE `ID`='3'"; 

$result = mysqli_query($conn, $sql); 

$sql = "SELECT `Played` FROM `Games` WHERE `ID` = 3"; 

$result = mysqli_query($conn, $sql); 

[...] 
+1

您忘了'WHERE'ID'='3'' – TiiJ7

+0

就是这样,非常感谢! – Tim

+1

它应该反映出答案^'WHERE'ID'='3'“;'为'WHERE ID ='3'”;'或者使用滴答声''''而不是普通引号''',请编辑您的答案。不正确的标识符http://dev.mysql.com/doc/refman/5.0/en/identifier-qualifiers.html –