jqueryのAJAX呼び出しは、関数呼び出しに取り組んでいない

Question

function GetSelectedTextValue() { 
    var selMenu = document.getElementById("selMenu"); 
    var selMenuValue = selMenu.value; 
    var selOptions=document.getElementById("selOptions"); 
    var selOptionsValue = selOptions.value; 
    var UserInput=$("#txtClaimNumber").val(); 
    if(selMenuValue=="Claims_Related" && selOptionsValue=="1") { 
     Claim_related_api(UserInput); 
    } 
} 

function Claim_related_api(Claim_number) { 
    var model = { PolicyNumber: "236426891", ClaimNumber: Claim_number }; 

    $.ajax({ 
     type: "Post", 
     data: JSON.stringify(model),     
     url: 'http://asplb-dev-00433/api/ClaimDetailsByClaim', 
     dataType : "json", 
     contentType: "application/json" 

    }).done(function (res) {    
     alert(res[0].CLAIM_ID); 

    }); 
} 

Answer 1

あなたは `Claim_related_api`を関数を呼び出す方法をあなたはこの

function Claim_related_api(Claim_number) { 
    var model = { PolicyNumber: "236426891", ClaimNumber: Claim_number }; 



    $.ajax({ 
     type: "Post", 
     data: JSON.stringify(model),     
     url: 'http://asplb-dev-00433/api/ClaimDetailsByClaim', 
     dataType : "json", 
     contentType: "application/json", 
     success: function (response) { 
      console.log(response); 

     }, 
     error: function() { 
      console.log(error); 
     }, 
    }); 

}