zoukankan      html  css  js  c++  java
  • 四元表达式

    // 展示安全问题详情
            ShowSecurityQues (item) {
                const itemCode = item.label
                this.detailsTitle = itemCode
                const SecurityNum = this.SecurityInfo.findIndex((item) => {
                    return item.checkCode === itemCode
                })
                this.work_content_info = [...this.security_content_info]
                this.work_content_info.forEach(item => {
                    const SecurityProp = this.SecurityInfo[SecurityNum][item.prop]
                    if (item.prop === 'state') {
                        item.info = (SecurityProp === 'back' ? '退回' : (SecurityProp === 'end' ? '通过' : '审批中'))
                this.activeColor = (SecurityProp === 'back' ? '#FF1A1A' : (SecurityProp === 'end' ? '#00D390' : '#FA6400')) } else if (item.prop === 'imgUrl') { this.imageUrl = [] let imgUrls if (SecurityProp !== '') { imgUrls = SecurityProp.split(',') } if (imgUrls && imgUrls.length > 0) { imgUrls.forEach(item => { this.imageUrl.push(window.g.baseUrl + '/downLoad1/' + item + '/' + getStore('bcToken')) }) } else { this.imageUrl = [] } } else if (item.prop === 'problemLevel') { item.info = (SecurityProp === '0' ? '一般' : (SecurityProp === '1' ? '重大' : '严重')) } else { item.info = SecurityProp } }) },

    参考:https://www.cnblogs.com/szqtiger/p/12108371.html

  • 相关阅读:
    P1338 末日的传说
    P1364医院设置
    线程
    进程通信
    CentOS设置中文
    C++快读讲解
    迭代加深搜索
    P1118 [USACO06FEB]Backward Digit Sums G/S
    N皇后问题
    RMQ区间最值查询
  • 原文地址:https://www.cnblogs.com/jvziking/p/15124861.html
Copyright © 2011-2022 走看看