Question

我已将问题缩小到这段代码：

<?php

$mysql_host = "mysqlXX.000webhost.com";
$mysql_user = "a4935911_******";
$mysql_password = "******";

$con = mysql_connect($mysql_host, $mysql_user, $mysql_password) or die(mysql_error());

$db = mysql_select_db('a4935911_******', $con) or die(mysql_error());

$getuname = mysqli_fetch_assoc(mysqli_query($con, "SELECT username FROM users WHERE id = 4"));
$username = $getuname ['username'];
echo "Username = " .$username;

?>

我在两页上有这个。第一个工作正常，输出'admin'按预期输出，而第二个给我两个错误：

警告：mysqli_query（）要求参数1为mysqli，第11行/home/a4935911/public_html/requestUser.php中给出的资源

警告：mysqli_fetch_assoc（）期望参数1为mysqli_result，在第11行的/home/a4935911/public_html/requestUser.php中给出为null

这怎么只会导致其中一个页面出现问题？有什么帮助吗？

更新

<?php
$con=mysqli_connect("mysqlXX.000webhost.com","a4935911_XXX","XXXX","a4935911_XXX");
$id = $_SESSION['uid'];
$getuname = mysqli_fetch_assoc(mysqli_query($con, "SELECT username FROM users WHERE id =   $id"));
$username = $getuname ['username'];
echo "Username = " .$username;
?>

错误：

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in /home/a4935911/public_html/requestUser.php on line 4

PHP是无情的。这次我错过了什么？你可以看到我对PHP很新。

Answer 1

您正尝试将mysqli用于mysql_ *函数。这是两件截然不同的事情。看看mysqli_connect功能。

启动它的错误是：

警告：mysqli_query（）要求参数1为mysqli，resource 给定

这意味着mysqli_query函数期望将mysqli链接作为其第一个参数。相反，它收到一个资源，因为mysql_connect将返回一个MySQL链接标识符资源变量。来自manual：

的mysqli示例

<?php
$mysqli = new mysqli('localhost', 'my_user', 'my_password', 'my_db');

/*
 * This is the "official" OO way to do it,
 * BUT $connect_error was broken until PHP 5.2.9 and 5.3.0.
 */
if ($mysqli->connect_error) {
    die('Connect Error (' . $mysqli->connect_errno . ') '
            . $mysqli->connect_error);
}

/*
 * Use this instead of $connect_error if you need to ensure
 * compatibility with PHP versions prior to 5.2.9 and 5.3.0.
 */
if (mysqli_connect_error()) {
    die('Connect Error (' . mysqli_connect_errno() . ') '
            . mysqli_connect_error());
}

echo 'Success... ' . $mysqli->host_info . "\n";

$mysqli->close();
?>

Answer 2

试试这个：

$mysql_host = "mysqlXX.000webhost.com";
$mysql_user = "a4935911_******";
$mysql_password = "******";

$con = new mysqli($mysql_host, $mysql_user, $mysql_password,'database_name');
if (mysqli_connect_errno()) die('Could not connect: ' . mysqli_connect_error());

$getuname = mysqli_fetch_assoc(mysqli_query($con, "SELECT username FROM users WHERE id = 4"));
$username = $getuname ['username'];
echo "Username = " .$username;

?>

<强>更新

<?php
$con=mysqli_connect("mysqlXX.000webhost.com","a4935911_XXX","XXXX","a4935911_XXX");
$id = $_SESSION['uid'];
$result=$con->query("SELECT username FROM users WHERE id =   $id");
$getuname = mysqli_fetch_array($result);
$username = $getuname ['username'];
echo "Username = " .$username;
?>

PHP代码在一个页面上获取错误并在另一个页面上工作

2 个答案: