我已经安装了客户端,并使用composer dump autoload进行了更新,但我仍然遇到了同样的错误.在通过composer安装后,需要在项目目录中使用guzzlehttp/guzzle:~6.0.
$client = new GuzzleHttp\Client();
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为什么它没有工作,为什么它甚至引用了错误的目录?
我尝试使用Proguard将我的代码导出到.apk文件,但是收到了很多警告.我尝试了很多解决方案,但无法修复它.有任何想法吗?
[2012-12-02 22:24:05 - com.crossrt.showtime.Main] Warning: android.support.v4.widget.ScrollerCompatIcs: can't find referenced method 'float getCurrVelocity()' in class android.widget.Scroller
[2012-12-02 22:24:05 - com.crossrt.showtime.Main] Warning: android.support.v4.widget.SearchViewCompatHoneycomb$1: can't find referenced class android.widget.SearchView$OnQueryTextListener
[2012-12-02 22:24:05 - com.crossrt.showtime.Main] Warning: android.support.v4.widget.SearchViewCompatHoneycomb$1: can't find referenced class android.widget.SearchView
[2012-12-02 22:24:05 - com.crossrt.showtime.Main] Warning: there were 242 unresolved references to classes or interfaces.
[2012-12-02 22:24:05 - com.crossrt.showtime.Main] You may need to specify additional library jars (using '-libraryjars').
[2012-12-02 22:24:05 - com.crossrt.showtime.Main] Warning: there were 47 unresolved references to program class members. …Run Code Online (Sandbox Code Playgroud) 我试图按照此处的说明操作,尝试将 Spring Boot 应用程序作为 init.d 服务运行,但未能成功。
我如上所述创建了完全可执行的 jar (myapp.jar),并创建了指向/etc/init.d/myapp的符号链接当我运行java -jar myapp.jar 时,我可以看到应用程序成功启动。
但是当我尝试使用
service myapp status 它说 Not running (process not found)
service myapp start 它说 Failed to start
文档说“假设您在 /var/myapp 中安装了 Spring Boot 应用程序”,我不太明白这一点。我将可执行 jar(通过 Jenkins)复制到 /var/myapp。所以它只包含一个 jar 文件。这会造成问题吗?
任何建议表示赞赏。
环境:
springBootVersion = '1.3.2.RELEASE'
JDK6 (yes it is)
Ubuntu 12.04
Run Code Online (Sandbox Code Playgroud) 我正在尝试使用PHP composer安装Magento 2.1.
http://idroot.net/tutorials/how-to-install-php-composer-on-centos-7/
我发现上面的url为Centos 7安装了PHP编写器.
在第2步失败了:
curl -sS https://getcomposer.org/installer | PHP
这是我得到的错误
"卷曲:(6)无法解析主机:php;名称或服务未知
谢谢你的帮助!
我正在尝试构建一个简单的PHP记录插入,但我似乎无法让它工作,它一直说我的语法有错误.
这不会出现在实时生产服务器上,这仅限于内部(我的使用),因此注入/攻击不是问题.
代码:
<?php
$servername = "server";
$username = "un";
$password = "p";
$dbname = "db";
$title = $_POST['title'];
$firstname = $_POST['firstname'];
$surname = $_POST['surname'];
$gender = $_POST['gender'];
$address = $_POST['address'];
$dob = $_POST['dob'];
$evetelephone = $_POST['eve-telephone'];
$daytelephone = $_POST['day-telephone'];
$occupation = $_POST['occupation'];
$email = $_POST['email'];
$hearabout = $_POST['hear-about'];
$consent = $_POST['consent'];
$readrules = $_POST['readrules'];
$injury = $_POST['injury'];
$questions = $_POST['questions'];
$gymrisk = $_POST['gymrisk'];
$rules = $_POST['rules'];
$remove = $_POST['remove'];
$fullname = $_POST['fullname'];
$signature = 'test';/*$_POST['signature'];*/
try {
$conn …Run Code Online (Sandbox Code Playgroud) 我目前正在运行多个mysql查询但想要运行一个并循环遍历它.我怎样才能改善这个?
$query = mysql_query("SELECT productid, code, size, quantity, sizechart, rprice, price, weight FROM inventory WHERE code = '$productcode1'");
while ($rows1 = mysql_fetch_array($query))
{
$productids1 = $rows1['productid'];
$codes1 = $rows1['code'];
$sizes1 = ucwords($rows1['size']);
$quantitys1 = $rows1['quantity'];
$sizechart = $rows1['sizechart'];
$rprice = $rows1['rprice'];
$sprice = $rows1['price'];
$dweight = $rows1['weight'];
}
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然后每个查询都进行相同但$ productcode2和$ productcode3等.
php ×4
composer-php ×2
mysql ×2
android ×1
centos ×1
centos7 ×1
guzzlehttp ×1
java ×1
laravel ×1
laravel-5 ×1
linux ×1
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proguard ×1
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