Php 这个代码怎么了?
我写了这篇文章,但它不起作用,找不到错误。 此服务器端代码获取变量$cpu&$display,并在数据库的select中使用它。当变量不重要时,将发送*Php 这个代码怎么了?,php,mysql,sql,Php,Mysql,Sql,我写了这篇文章,但它不起作用,找不到错误。 此服务器端代码获取变量$cpu&$display,并在数据库的select中使用它。当变量不重要时,将发送* <?php if (isset($_REQUEST['action'])) { $action = $_REQUEST['action']; } else { echo "Invalid Data"; exit; } if ($action == "read") { readData(); } func
<?php
if (isset($_REQUEST['action']))
{
$action = $_REQUEST['action'];
}
else
{
echo "Invalid Data";
exit;
}
if ($action == "read")
{
readData();
}
function connectToDatabase()
{
$connection = mysqli_connect("localhost", "root", "", "project_pro");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
return $connection;
}
function readData()
{
$connection = connectToDatabase();
$cpu = $_REQUEST['cpu'];
$display = $_REQUEST['display'];
PHP中的连接运算符是。不是+。因此,将+=改为.=。好吧,由于缺乏适当的缩进,阅读起来非常困难,例如…:p+=是javascript,不是php!:。php中的连接是。=+当他从未执行查询时,他试图获取$row=mysqli\u fetch\u array$result。。
$sql = "Select * From phones WHERE";
if ($cpu == "*")
{
}
else
{
$sql+= " phone_cpu='$cpu'";
}
if ($display == "*")
{
}
else
{
$sql+= " AND phone_display='$display'";
}
$output = array();
while ($row = mysqli_fetch_array($result))
{
$record = array();
$record['phone_id'] = $row['phone_id'];
$record['phone_cpu'] = $row['phone_cpu'];
$output[] = $record;
}
echo json_encode($output);
mysqli_close($connection);
}