我的html文件中有以下代码：
<Form Name ="form1" Method ="POST" ACTION = "login.php">
<label>Username: </label>
<input type="text" name="pat_username" id="username" required="required" placeholder="username"/><br/><br />
<label>Password: </label>
<input type="password" name="pat_password" id="password" required="required" placeholder="password"/><br/><br />
<input type="submit" value=" Submit " name="submit"/><br />
</form>
以及login.php中的以下内容：
<?php
if(isset($_POST["submit"])){
$servername = "localhost";
$username = "root";
$password = "mysql123";
$dbname = "hospital";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
//Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$newUsername = mysqli_real_escape_string($conn, $_POST['pat_username']);
$newPassword = mysqli_real_escape_string($conn, $_POST['pat_password']);
$result = $conn->query("SELECT * FROM patients WHERE patient_username ='$newUsername' AND patient_password='$newPassword'");
if (mysqli_num_rows($result)) {
echo "SELECT * FROM patients WHERE patient_username='$newUsername' AND patient_password='$newUsername'";
//echo "login successfull";
}
else
{
echo "login unsuccessfull, please try again";
}
$conn->close();
}
?>
php文件可以检查＆＃39; patient_username＆＃39;和＆＃39; patient_password＆＃39;在＃39;患者中＆＃39;桌子位于医院＆＃39;数据库。当用户输入用户名和密码并点击提交时，“＆＃39;患者＆＃39;然后应检查表格，以确保输入的用户名和密码符合＆＃39; patient_username＆＃39;和＆＃39; patient_password＆＃39;在＃39;患者中＆＃39;表。输入用户名和密码并单击“提交”后，将显示以下错误：
Fatal error: Call to a member function mysqli_num_rows() on boolean in C:\xampp\htdocs\xampp\patientLogin\www\login.php on line 23
第27行：if（$ result-＆gt; mysqli_num_rows（））{
我最近才开始使用PHP和mysqli进行项目，我感觉很失落。任何人都可以告诉我这个吗？
解决了！感谢大家的帮助。我已经更新了上面的PHP代码^
答案 0 :(得分：0)
替换
<content src="index.html" />
<access origin="*" />
<allow-navigation href="*" />
<allow-intent href="*.firebaseio.com" />
<allow-intent href="auth.firebase.com" />
<preference name="webviewbounce" value="false" />
<preference name="UIWebViewBounce" value="false" />
<preference name="DisallowOverscroll" value="true" />
<preference name="BackupWebStorage" value="none" />
<preference name="SplashScreen" value="screen" />
<preference name="SplashScreenDelay" value="3000" />
<preference name="AutoHideSplashScreen" value="false" />
<preference name="ShowSplashScreenSpinner" value="false" />
<preference name="CordovaWebViewEngine" value="CDVWKWebViewEngine" />
<feature name="StatusBar">
<param name="ios-package" onload="true" value="CDVStatusBar" />
</feature>
<feature name="SocialSharing">
<param name="id" value="nl.x-services.plugins.socialsharing" />
<param name="url" value="https://github.com/EddyVerbruggen/SocialSharing-PhoneGap-Plugin.git" />
</feature>
<feature name="InAppBrowser">
<param name="id" value="cordova-plugin-inappbrowser" />
<param name="url" value="https://git-wip-us.apache.org/repos/asf/cordova-plugin-inappbrowser.git" />
</feature>
<feature name="InAppBrowser">
<param name="ios-package" value="CDVInAppBrowser" />
</feature>
<feature name="InAppBrowser">
<param name="android-package" value="org.apache.cordova.InAppBrowser" />
</feature>
<feature name="CDVWKWebViewEngine">
<param name="ios-package" value="CDVWKWebViewEngine" />
</feature>
<platform name="ios">
<icon height="57" src="resources/ios/icon/icon.png" width="57" />
<icon height="114" src="resources/ios/icon/icon@2x.png" width="114" />
<icon height="40" src="resources/ios/icon/icon-40.png" width="40" />
<icon height="80" src="resources/ios/icon/icon-40@2x.png" width="80" />
<icon height="50" src="resources/ios/icon/icon-50.png" width="50" />
<icon height="100" src="resources/ios/icon/icon-50@2x.png" width="100" />
<icon height="60" src="resources/ios/icon/icon-60.png" width="60" />
<icon height="120" src="resources/ios/icon/icon-60@2x.png" width="120" />
<icon height="180" src="resources/ios/icon/icon-60@3x.png" width="180" />
<icon height="72" src="resources/ios/icon/icon-72.png" width="72" />
<icon height="144" src="resources/ios/icon/icon-72@2x.png" width="144" />
<icon height="76" src="resources/ios/icon/icon-76.png" width="76" />
<icon height="152" src="resources/ios/icon/icon-76@2x.png" width="152" />
<icon height="29" src="resources/ios/icon/icon-small.png" width="29" />
<icon height="58" src="resources/ios/icon/icon-small@2x.png" width="58" />
<icon height="87" src="resources/ios/icon/icon-small@3x.png" width="87" />
<splash height="1136" src="resources/ios/splash/Default-568h@2x~iphone.png" width="640" />
<splash height="1334" src="resources/ios/splash/Default-667h.png" width="750" />
<splash height="2208" src="resources/ios/splash/Default-736h.png" width="1242" />
<splash height="1242" src="resources/ios/splash/Default-Landscape-736h.png" width="2208" />
<splash height="1536" src="resources/ios/splash/Default-Landscape@2x~ipad.png" width="2048" />
<splash height="768" src="resources/ios/splash/Default-Landscape~ipad.png" width="1024" />
<splash height="2048" src="resources/ios/splash/Default-Portrait@2x~ipad.png" width="1536" />
<splash height="1024" src="resources/ios/splash/Default-Portrait~ipad.png" width="768" />
<splash height="960" src="resources/ios/splash/Default@2x~iphone.png" width="640" />
<splash height="480" src="resources/ios/splash/Default~iphone.png" width="320" />
</platform>
<icon src="resources/ios/icon/icon-small@3x.png" />
<plugin name="org.apache.cordova.splashscreen" spec="~1.0.0" />
<plugin name="ionic-plugin-keyboard" spec="~1.0.8" />
<plugin name="cordova-plugin-camera" spec="~2.1.0" />
<plugin name="cordova-plugin-whitelist" spec="~1.2.1" />
<plugin name="org.apache.cordova.device" spec="~0.3.0" />
<plugin name="phonegap-plugin-push" spec="~1.5.3" />
<plugin name="cordova-plugin-badge" spec="~0.7.2" />
<plugin name="phonegap-plugin-barcodescanner" spec="https://github.com/phonegap/phonegap-plugin-barcodescanner.git" />
<plugin name="cordova-plugin-nativeaudio" spec="~3.0.6" />
与
$result = mysqli_query($connect, "SELECT * FROM patients WHERE patient_username='".$username."' AND patient_password='".$password."' AND activate='0'");
因为您在
中为 $result = mysqli_query($conn, "SELECT * FROM patients WHERE patient_username='".$username."' AND patient_password='".$password."' AND activate='0'");
而非$conn分配了类引用
$connect还有一个混合了oops和程序格式的东西;
使用类// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
oops格式连接并使用过程格式mysqli进行查询;
现在你可以改变查询的方式来oops喜欢这个，休息很好;
mysqli_query您应该使用prepared statement来阻止sql injection;
答案 1 :(得分：0)
由于您使用 MySQLi扩展（“i”代表改进）：面向对象，您需要为 mysqli_real_escape_string（）提供两个参数。
1）连接的第一个参数
2） String to Esacpe
的第二个参数您需要更改以下行：
$username = $conn->mysqli_real_escape_string($_POST['pat_username']);
$password = $conn->mysqli_real_escape_string($_POST['pat_password']);
由于您使用的是 MySQLi（面向对象），它是 MySQL 的后续版本，因此它们对其进行了一些改进，就像其他改进一样（您已经在使用你的代码。就像执行查询时一样，你也可以在这里给出第二个连接参数，你需要给它。）
更新1：
$result = $conn->query("SELECT * FROM patients WHERE patient_username ='$username' AND patient_password='$password' AND activate='0'");
// you do not need a dot (concatenation)
更新2：
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "hospital";
现在，$password和$username目前的价值如上所述。
$username = mysqli_real_escape_string($conn, $_POST['pat_username']);
$password = mysqli_real_escape_string($conn, $_POST['pat_password']);
现在再次使用相同的变量名称来获取用户获得的表单。因此，在执行以下查询时：
$result = $conn->query("SELECT * FROM patients WHERE patient_username='".$username."' AND patient_password='".$password."' AND activate='0'");
连接失败，因为您要按表单值更新变量$password和$username。
所以只需更改连接的变量或用于从表单获取值的变量。
我希望这会对你有所帮助。
答案 2 :(得分：-1)
通过仅使用OOP变体，您错过了存储结果。
替换：
$conn->query("SELECT * FROM patients WHERE patient_username='".$username."' AND patient_password='".$password."' AND activate='0'");
使用：
$result = $conn->query("SELECT * FROM patients WHERE patient_username='".$username."' AND patient_password='".$password."' AND activate='0'");
或者与上述程序功能相同。
（同时打开错误显示error_reporting(E_ALL);和ini_set('display_errors', 1);，并确保在服务器上启用了mysqli扩展。顺便说一下，我建议PDO用于数据库连接）