所以我理解并知道ajax是如何工作的,因为我之前已经使用过它.但是如何从PHP脚本中检索
JSON数据呢？
我试图回应json_encode结果(这是有效的),但我得到一个例外：未捕获(在承诺中)：意外的令牌<在Chrome中的位置0错误的JSON中. 错误的行：
search(): any
{
this.http.get('app/PHP/search.PHP').toPromise()
.then(response => this.data = response.json())
.catch(this.handleError);
}
$json = array (
"age" => 5,"bob" => "Lee",);
echo json_encode($json);
解决方法
你为什么不使用observables ..我在我的虚拟项目中使用了PHP作为后端.这是它的代码.
ngOnInit() {
let body=Path+'single.PHP'+'?id=' + this.productid;
console.log(body);
this._postservice.postregister(body)
.subscribe( data => {
this.outputs=data;
// console.log(this.outputs);
},error => console.log("Error HTTP Post Service"),() => console.log("Job Done Post !") );
}
$faillogin=array("error"=>1,"data"=>"no data found");
$successreturn[]=array(
"productid"=>"any","productname"=>"any","productprice"=>"any","productdescription"=>"any","productprimaryimg"=>"any","otherimage"=>"any","rating"=>"any");
// Create connection id,name,price,description,primary_image,other_image,rating
$productid = $_GET["id"];
$sql="SELECT * FROM product_list where id='$productid'";
$result = MysqLi_query($conn,$sql);
$count = MysqLi_num_rows($result);
$value=0;
while($line = MysqLi_fetch_assoc($result))
{
$successreturn[$value]['productid']=$line['id'];
$successreturn[$value]['productname']=$line['name'];
$successreturn[$value]['productprice']=$line['price'];
$successreturn[$value]['productdescription']=$line['description'];
$successreturn[$value]['productprimaryimg']=$line['primary_image'];
$successreturn[$value]['otherimage']=$line['other_image'];
$successreturn[$value]['rating']=$line['rating'];
$value++;
}
echo json_encode($successreturn);
MysqLi_close($conn);
