Question

我想用MySQL输出MySQL数据库中的一些分数。在我的select查询之后，我尝试回显该值，但我得到的只是RESOURCE ID 10＃。这是我的代码：

$rank = mysql_query("SELECT 1+COUNT(*) FROM `class_ranking` WHERE overall_scores > (SELECT overall_scores FROM `class_ranking` WHERE ref_id='$id')");
While ($rk = mysql_fetch_array($rank)){
    echo $rk['overall_scores'];
}

我需要你的帮助。感谢。

Answer 1

以这种方式写1+COUNT(*) as overall_scores

$rank = mysql_query("SELECT 1+COUNT(*) as overall_scores FROM `class_ranking` WHERE overall_scores > (SELECT overall_scores FROM `class_ranking` WHERE ref_id='$id')");
While ($rk = mysql_fetch_array($rank)){
    echo $rk['overall_scores'];
}

Answer 2

设置计数的别名：

apply plugin: 'com.android.application' android { compileSdkVersion 26 defaultConfig { applicationId "tomestore.wixsite.com.tome" minSdkVersion 19 targetSdkVersion 26 versionCode 1 versionName "1.0" testInstrumentationRunner "android.support.test.runner.AndroidJUnitRunner" } buildTypes { release { minifyEnabled false proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro' } } } dependencies { implementation fileTree(dir: 'libs', include: ['*.jar']) implementation 'com.android.support:appcompat-v7:26.1.0' implementation 'com.android.support.constraint:constraint-layout:1.0.2' implementation 'com.google.firebase:firebase-storage:11.0.4' implementation 'com.google.firebase:firebase-auth:11.0.4' implementation 'com.google.firebase:firebase-database:11.0.4' implementation 'com.google.firebase:firebase-messaging:11.0.4' implementation 'com.android.support:recyclerview-v7:26.1.0' testImplementation 'junit:junit:4.12' androidTestImplementation 'com.android.support.test:runner:1.0.1' androidTestImplementation 'com.android.support.test.espresso:espresso-core:3.0.1' implementation 'com.android.support:design:26.1.0' implementation 'com.android.support:support-v4:26.1.0' // RecyclerView compile 'com.android.support:recyclerview-v7:26.1.0' // CardView compile 'com.android.support:cardview-v7:26.1.0' } apply plugin: 'com.google.gms.google-services'

Answer 3

首先，SELECT 1+COUNT(*)将返回类似

的内容

+------------+
| 1+count(*) |
+------------+
|          4 |
+------------+

因此，这不起作用

While ($rk = mysql_fetch_array($rank)){
    echo $rk['overall_scores'];
}

因为没有overall_scores。您可能必须将其别名为SELECT 1+COUNT(*) AS overall_scores FROM...并执行

$result=mysql_query("SELECT 1+COUNT(*) AS overall_scores FROM ...");
$data=mysql_fetch_assoc($result);
echo $data['overall_scores']; // Or $data['1+COUNT(*)'] if you did not alias it

Answer 4

你需要改变mysql_fetch_array（）而不是mysql_fetch_assoc（）看起来像这样

$rank = mysql_query("SELECT 1+COUNT(*) AS overall_scores FROM `class_ranking` WHERE overall_scores > (SELECT overall_scores FROM `class_ranking` WHERE ref_id='$id')");
$finalArr = [];
While ($rk = mysql_fetch_assoc($rank)){
    $finalArr[] = $rk['overall_scores'];
}
echo '<pre>';
print_r($finalArr);

在php中输出资源变量的值

4 个答案: