2016-08-02 100 views
0

嘿傢伙我試圖創建搜索與反應原生的後端API,我必須將輸入到TextInput的單詞傳遞到URL。我不知道我是否正確地做了這件事,任何機構能否幫助我糾正如何將在textinput中輸入的值傳遞給url?

這是代碼。

this.state = { 
     search: "", 
    } 

async onSearchPressed() { 
    try { 
     let response = await fetch("http://www.endpoints.com/search/{this.state.search}", { 
     method: "GET", 
     headers: { 
      'Accept': 'application/json', 
      'Content-Type': 'application/json' 
     }, 
     }); 


     render =() => { 
    let fields = [ 
      {ref: 'search', placeholder: 'search', keyboardType:'default',secureTextEntry: false},]; 
    return (
     <TextInput 
      {...fields[0]} 
      onChangeText={(val) => this.setState({search: val})} 
      value={this.state.search} 
     /> 

     <TouchableOpacity onPress={this.onSearchPressed.bind(this)} /> 
+0

你檢查在'onSearchPressed()'函數生成的網址是什麼? – Sriraman

+0

我該如何檢查? – atif

+0

檢查我的答案。 – Sriraman

回答

1

看起來,它採取的{this.state.search}爲字符串。

變化

let response = await fetch("http://www.endpoints.com/search/{this.state.search}", { 

let response = await fetch("http://www.endpoints.com/search/"+this.state.search, { 
+0

感謝sriraman整頓 – atif