如何将textinput中输入的值传递到url?
嘿,伙计们,我正在尝试用react native中的后端API创建搜索,我必须将输入到TextInput中的单词传递到url。我不知道我做得对不对,有没有人能帮我改正 这是代码如何将textinput中输入的值传递到url?,url,search,react-native,backend,textinput,Url,Search,React Native,Backend,Textinput,嘿,伙计们,我正在尝试用react native中的后端API创建搜索,我必须将输入到TextInput中的单词传递到url。我不知道我做得对不对,有没有人能帮我改正 这是代码 this.state = { search: "", } async onSearchPressed() { try { let response = await fetch("http://www.endpoints.com/search/{this.state.search}
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)} />
this.state={
搜索:“,
}
异步onSearchPressed(){
试一试{
let response=等待获取(“http://www.endpoints.com/search/{this.state.search}“{
方法:“获取”,
标题:{
“接受”:“应用程序/json”,
“内容类型”:“应用程序/json”
},
});
渲染=()=>{
设字段=[
{ref:'search',占位符:'search',键盘类型:'default',secureTextEntry:false},];
返回(
this.setState({search:val})}
值={this.state.search}
/>
您尝试使用的是
应该是这样的:
fetch(`http://www.endpoints.com/search/${this.state.search}`
看起来,它将
{this.state.search}
作为字符串
改变
let response = await fetch("http://www.endpoints.com/search/{this.state.search}", {
到
您是否在
onSearchPressed()
函数中检查了生成的URL?如何检查?
let response = await fetch("http://www.endpoints.com/search/"+this.state.search, {