首页 > 其他分享 >I Will Boast In Christ Lyric

I Will Boast In Christ Lyric

时间:2023-06-25 13:32:32浏览次数:32  
标签:Christ but Jesus Will Lyric blood Nothing my

I Will Boast In Christ Lyric

(Come on, let's make yourself a person tonight)
All I have because of Jesus
All this promise
Won for me
When He paid the highest ransom
Once for always
For my freedom
I will boast in Christ alone
His righteousness
Not my own
I will cling to Christ my hope
His mercy reigns
Now and forever
Love will never lose its power
All my failures could not erase
Now I walk within Your favour
Grace unending
My salvation

(Come one, let's declare this tonight)
I will boast in Christ alone
His righteousness
And Not my own
I will cling to Christ my hope
His mercy reigns
Now and forever
What can wash away my sin
Nothing but the blood of Jesus
What can make me whole again
Nothing but the blood of Jesus

(Come on, let's fill this place with singing)
What can wash away my sin
Nothing but the blood of Jesus
What can make me whole again
Nothing but the blood of Jesus
Nothing but the blood of Jesus
(I will boast)
I will boast in Christ alone
His righteousness
And Not my own
I will cling to Christ my hope
His mercy reigns
Now and forever
What can wash away my sin
Nothing but the blood of Jesus
What can make me whole again
Nothing but the blood of Jesus

(O precious)
O precious is the flow
That makes me white as snow
No other fount I know
Nothing but the blood of Jesus
Nothing but the blood of Jesus


欢迎关注公-众-号【TaonyDaily】、留言、评论,一起学习。

Don’t reinvent the wheel, library code is there to help.

刘俊涛的博客


若有帮助到您,欢迎点赞、转发、支持,您的支持是对我坚持最好的肯定(_)

你要保守你心,胜过保守一切。

作者:刘俊涛的博客,


标签:Christ,but,Jesus,Will,Lyric,blood,Nothing,my
From: https://blog.51cto.com/love/6545182

相关文章

  • 8月最新-《可解释机器学习-Christoph Molnar》-新书分享
        机器学习在改进产品、过程和研究方面拥有很大的潜力。但是机器学习模型预测的结果通常是不可解释的,这也是机器学习技术最大不足。本书主要讲解如何搭建机器学习模型,并使他们的预测结果是可解释的。 (文末附本书免费下载地址)    本书首先讲解可解释性的基本概念,然后讲......
  • CF248B Chilly Willy 题解
    CF248BChillyWilly解题过程经过简单思考,这道题肯定是由规律可循,因为\(n\le10^5\),只有高精度能存下。下面是暴力程序对\(n\)为\(1\)到\(13\)时的答案进行求解(\(11\)到\(13\)超出int范围了)。发现\(n\)为\(1\)或\(2\)时,他们的答案为\(-1\)。接着来分析......
  • Windows系统管理大师、畅销书作者William R.Stanek的又一经典力作
    内容简介:本书是Windows系统管理大师、畅销书作者WilliamR.Stanek的又一经典力作,结构清晰、讲解透彻、方便实用。本书的独到之处在于,它从管理与技术相结合的角度,详细阐述了如何通过命令行来有效管理Windows操作系统。而且,作者从大多数系统管理员的日常工作需求出发,将全书分为日志管......
  • php解决 mysql_connect(): The mysql extension is deprecated and will be removed i
    Themysqlextensionisdeprecatedandwillberemovedinthefuture:usemysq翻译:mysql_connect这个模块将在未来弃用,请你使用mysqli或者PDO来替代。解决方法:打开php.ini配置文件把display_errors=On改为display_errors=Off改完之后重启服务就可以了。  ......
  • Custom directive is missing corresponding SSR transform and will be ignored
    背景最近在给业务组件库集成指令库,将各个项目中常用的指令如一键复制、元素和弹窗拖拽等封装到一起,进行统一发版维护。业务组件库项目架构采用的是pnpm+vite+vue3+vitepress,其中vitepress主要做组件库文档站点同时展示可交互的组件。问题开发运行时指令库demo没有问题,构建编译......
  • 报错:resolution will not be reattempted until the update interval of XXX has elap
     含义:在XXX的更新间隔过去或强制更新之前,不会重新尝试解析。如果你去本地的maven仓库,你会发现,其只有lastUpdate结尾的文件,没有jar包。这个时候,你无论怎么点击IDEA中的ReimportsAllMavenProjects都是没有用的。原因上面也说了,要么等更新时间过去,要么强制更新。maven的默认......
  • UIViewController生命周期方法viewDidLoad、viewWillAppear和viewDidAppear
    UIViewController生命周期方法viewDidLoad、viewWillAppear和viewDidAppear 这3个方法执行顺序为:viewDidLoad-》viewWillAppear-》viewDidAppear  viewDidLoadCalledaftertheviewhasbeenloaded.Forviewcontrollerscreatedincode,thisisafter-loadView.Forviewc......
  • 关于self.view.window与viewDidLoad、viewWillAppear、viewDidAppear
    关于self.view.window与viewDidLoad、viewWillAppear、viewDidAppear 在进入一个界面(UIViewController),如果要在进入的时候使用self.view.window,那么必须在将使用代码放在viewDidAppear方法中,而viewDidLoad、viewWillAppear中self.view.window.frame为0{{0,0},{0,0}}。 -(void)......
  • Ruby实践—will_paginate实现分页
    开发环境:OS:WindowsXPRuby:Ruby1.9.1Rails:Rails2.3.5will_paginate:will_paginate2.3.11(在命令行中运行geminstallmislav-will_paginate--sourcehttp://gems.github.com )IDE:Rubymine2.0.1DB:mysql5.0.9 本例在上一个例子(Ruby实践—简单数据库操作)的基础上实现分页,利用......
  • [ABC166E] This Message Will Self-Destruct in 5s
    ThisMessageWillSelf-Destructin5sの传送门Solution首先看到\(j-i=A_i+A_j\)转换一下,\(i+a_i=j-a_j\)。接下来,对于每一个\(i\)(\(1\lei\len\)),用一个map存\(i-a_i\)的数量。最后枚举\(i\)(\(1\lei\len\)),每次将\(ans\)加上\(i+a_i\)在map里的数......