当existence=1时再使用这个方法

Question

我想当我选择已在库中的时候再完整的使用setUserInfo这个方法，如果我选择未在库中，我只需要setUserInfo方法中的第一行： this.addForm.worker = this.showUsername;不然就会报错，请问应该怎么修改呢？

<div class="grid-content bg-purple">
                    
                    <el-radio border size="mini" @change="existenceChange(existence)" v-model="existence" :label="1">
                      已在库中
                    </el-radio>
                    <el-radio border size="mini" @change="existenceChange(existence)" v-model="existence" :label="0">
                      未在库中
                    </el-radio>
                  
 
<el-row>
                <el-col :span="8">
                  <el-form-item label="物资数量" prop="numbers" v-show="existence === 1">
                    <el-input v-model="addForm.numbers"></el-input>
                  </el-form-item>
                </el-col>
              </el-row>
 
 
                <el-row>
                  <el-col :span="8">
                    <el-form-item label="物资名称" prop="itemname" v-show="existence === 0">
                      <el-input v-model="addForm.itemname"></el-input>
                    </el-form-item>
                  </el-col>
                </el-row>
 
                <el-row>
                  <el-col :span="8">
                    <el-form-item label="物资数量" prop="numbers" v-show="existence === 0">
                      <el-input v-model="addForm.numbers"></el-input>
                    </el-form-item>
                  </el-col>
                </el-row>
 
                <el-row>
                  <el-col :span="8">
                    <el-form-item label="物资单位" prop="unit" v-show="existence === 0">
                      <el-input v-model="addForm.unit"></el-input>
                    </el-form-item>
                  </el-col>
                </el-row>
 

existenceChange(existence) {
      if (existence === 1) {
      } else if (existence === 0) {
      }
    },


selectedItems() {
      return this.stockList.filter(item => this.$refs.table.selection.includes(item))
          .map(item => ({ name: item.itemname, unit: item.unit }));
    }
 
addIn() {
      this.$refs.addFormRef.validate(async valid => {
        if (!valid) return;
        this.setUserInfo();
        const {data: res} = await this.$http.post("addIn", this.addForm);
        if (res != "success") {
          return this.$message.error("入库失败~");
        }
        this.$message.success("入库成功~");
        this.addDialogVisible = false;
        this.getStockList();
      });
    },
    setUserInfo() {
      // 填入用户名
      this.addForm.worker = this.showUsername;
      this.addForm.itemname = this.selectedItems[0].name;
      this.addForm.unit = this.selectedItems[0].unit;
 
    },
 

Answer 1

参考GPT和自己的思路：根据你提供的代码和问题描述，如果你选择未在库中，只需要执行 setUserInfo 方法的第一行代码：this.addForm.worker = this.showUsername;，那么你可以在代码中加上一个条件判断，只有当 existence 的值为 1 时才执行 setUserInfo 方法的其他部分。具体代码修改如下：

addIn() {
  this.$refs.addFormRef.validate(async valid => {
    if (!valid) return;
    if (this.existence === 1) {
      this.setUserInfo();
    } else {
      this.addForm.worker = this.showUsername;
    }
    const {data: res} = await this.$http.post("addIn", this.addForm);
    if (res != "success") {
      return this.$message.error("入库失败~");
    }
    this.$message.success("入库成功~");
    this.addDialogVisible = false;
    this.getStockList();
  });
},

这样，当 existence 的值为 1 时，才会执行 setUserInfo 方法的其他部分，而当 existence 的值为 0 时，只会执行 setUserInfo 方法的第一行代码。