我正在尝试编写我的第一个基本PHP RESTful API - 我设法使用MAMP在本地机器上运行它。但是当我上传到托管服务器时,它不想工作。PHP mysqli_query将不会执行
下面的代码 - 我在里面添加了一些ECHO,以确保事情一直在进行中。看起来我们都很好,直到$result=mysqli_query
。
<?php
//header('Content-type:application/json');
// Connect to db
$con = mysqli_connect("HOSTNAME","USER","PASSWORD","DATABASE");
echo "Database: ";
// Check connection
if ($con->connect_error) {
die("Connection failed: " . $con->connect_error);
}
echo "Connected successfully";
echo "<br><br>";
// Get value from url
$bid = $_GET['bid'];
echo "BID: ";
echo $bid;
echo "<br><br>";
// Define Query
$sql = "SELECT id, bandname, members, bio, songlist FROM bands WHERE id='$bid'";
echo "SQL Query: ";
echo $sql;
echo "<br><br>";
// Run Query
$result = mysqli_query($con, $sql);
echo "Result: ";
print_r($result);
echo "<br><br>";
// Put Query result into array
$result_array = mysqli_fetch_array($result, MYSQLI_ASSOC);
echo "Result Array: ";
print_r($result_array);
echo "<br><br>";
// Encode array as JSON and output
echo "JSON: ";
echo json_encode($result_array);
?>
我输入网址是http://bandsly.com/api.php?bid=1,这是我收到的浏览器输出...
Database: Connected successfully
BID: 1
SQL Query: SELECT id, bandname, members, bio, songlist FROM bands WHERE id='1'
Result:
Result Array:
JSON: null
当我手动运行在数据库中查询SELECT id, bandname, members, bio, songlist FROM bands WHERE id='1'
(phpMyAdmin的) ,它工作正常,我得到1行返回正确的值。
(http://i.stack.imgur.com/d3ZPJ.png)
任何帮助,将最理解!!
*****编辑*****
OK,我想我找到了问题...我的“连接成功”编写不正确,并且总是回来看起来很不错。它看起来像修复后,我有数据库连接问题。
我打算去查看db连接设置并尝试修复它。
谢谢大家的帮助!
你可以用'mysql_fetch_assoc($结果);'$ – aldrin27
CON = mysqli_connect( “主机名”, “用户”, “密码”, “数据库”);这里一切都好吗? – M0rtiis
谢谢@ aldrin27 - 我会尽量保持它的精益。 –