Question

我是jQuery的新手，并在node.js上使用了一个使用此代码段的旧教程：

$(function () {    
    var roomId;

    $.ajax({
        type: "GET",
        url: "/api/rooms"
    }).success(function (rooms) { 
        roomId = rooms[0].id;
        getMessages();
        $.each(rooms, function (key, room) {
            var a = '<a href="#" data-room-id="' + room.id + '" class="room list-group-item">' + room.name + '</a>';
            $("#rooms").append(a);
        });

    });

    [...]       

});

但是我收到此错误

未捕获的TypeError：$ .ajax（...）。success不是函数

在}).success(function (rooms) {

我想知道这里有什么不对吗？

Answer 1

对ajax的调用应如下所示：

$.ajax({
    type: "GET",
    url: "/api/rooms",
    success: function (rooms) { 

    }
});

你没有方法链接成功函数，它是字典参数中的一个条目。

Answer 2

您的代码是正确的，没有问题

但是你可能会包含新的jquery库，它不允许使用.success（）方法

对于较新版本的jquery使用

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>

<script>
    $.ajax({
        type: "GET",
        url: "/api/rooms",
        success: function (rooms) {

        }
    });
</script>

如果您使用旧的jquery，.success（）方法将运行没有任何问题

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>

<script>
    $.ajax({

        url: "/api/rooms",
        method: "GET",
        data: {'datavar': datavalue}

    }).success(function (rooms) {

        console.log("successfully run ajax request..." + rooms);

    }).done(function () {

        console.log("I am from done function");

    }).fail(function () {

        console.log("I am from fail function.");

    }).always(function () {

        console.log("I am from always function");

    });
</script>

Answer 3

根据documentation

从jQuery 3.0开始，#include <stdio.h> #include <stdlib.h> #include <string.h> #include "bignum_math.h" /* * Returns true if the given char is a digit from 0 to 9 */ bool is_digit(char c) { return c >= '0' && c <= '9'; } /* * Returns true if lower alphabetic character */ bool is_lower_alphabetic(char c) { return c >= 'a' && c <= 'z'; } /* * Returns true if upper alphabetic character */ bool is_upper_alphabetic(char c) { return c >= 'A' && c <= 'Z'; } /* * Convert a string to an integer * returns 0 if it cannot be converted. */ int string_to_integer(char* input) { int result = 0; int length = strlen(input); int num_digits = length; int sign = 1; int i = 0; int factor = 1; if (input[0] == '-') { num_digits--; sign = -1; } for (i = 0; i < num_digits; i++, length--) { if (!is_digit(input[length-1])) { return 0; } if (i > 0) factor*=10; result += (input[length-1] - '0') * factor; } return sign * result; } /* * Returns true if the given base is valid. * that is: integers between 2 and 36 */ bool valid_base(int base) { if(!(base >= 2 && base <= 36)) { return false; } return true; } /* * TODO * Returns true if the given string (char array) is a valid input, * that is: digits 0-9, letters A-Z, a-z * and it should not violate the given base and should not handle negative numbers */ bool valid_input(char* input, int base) { /* * check for valid base and if negative */ if (!valid_base(base) || input[0]=='-') { return false; } else { int len = strlen(input); int i; for (i =0; i< len; i++){ /* * check if the input string is a digit/letter */ if (!(is_digit(input[i]) || is_lower_alphabetic(input[i]) || is_upper_alphabetic(input[i]))){ return false; } /* * if the int excesses the base? */ else if (is_digit(input[i])){ if (input[i]-'0'>=base){ //convert char to int and compare with the base return false; } } /* *or if the letter excesses the base? */ else if (is_lower_alphabetic(input[i])){ if (input[i]-'a'+10 >=base){ return false; } } else if (is_upper_alphabetic(input[i])){ if (input[i] - 'A' + 10 >=base) { return false; } } } return true; } } /* * converts from an array of characters (string) to an array of integers */ int* string_to_integer_array(char* str) { int* result; int i, str_offset = 0; result = malloc((strlen(str) + 1) * sizeof(int)); result[strlen(str)] = -1; for(i = str_offset; str[i] != '\0'; i++) { if(is_digit(str[i])) { result[i - str_offset] = str[i] - '0'; } else if (is_lower_alphabetic(str[i])) { result[i - str_offset] = str[i] - 'a' + 10; } else if (is_upper_alphabetic(str[i])) { result[i - str_offset] = str[i] - 'A' + 10; } else { printf("I don't know how got to this point!\n"); } } return result; } /* * finds the length of a bignum... * simply traverses the bignum until a negative number is found. */ int bignum_length(int* num) { int len = 0; while(num[len] >= 0) { len++; } return len; } /* * TODO * Prints out a bignum using digits and upper-case characters * Current behavior: prints integers * Expected behavior: prints characters */ void bignum_print(int* num) { int i; if(num == NULL) { return; } /* Handle negative numbers as you want * let the last digit be -2 if negative * */ i = bignum_length(num); if (num[i]==-2){ printf("-"); } /* Then, print each digit */ for(i = 0; num[i] >= 0; i++) { if (num[i]<=9){ printf("%d", num[i]); } else if (num[i]>9){ char digit = num[i]+'A'-10; printf("%c", digit); } } printf("\n"); } /* * Helper for reversing the result that we built backward. * see add(...) below */ void reverse(int* num) { int i, len = bignum_length(num); for(i = 0; i < len/2; i++) { int temp = num[i]; num[i] = num[len-i-1]; num[len-i-1] = temp; } } /* * used to add two numbers with the same sign * GIVEN FOR GUIDANCE */ int* add(int* input1, int* input2, int base) { int len1 = bignum_length(input1); int len2 = bignum_length(input2); int resultlength = ((len1 > len2)? len1 : len2) + 2; int* result = (int*) malloc (sizeof(int) * resultlength); int r = 0; int carry = 0; int sign = input1[len1]; len1--; len2--; while(len1 >= 0 || len2 >= 0) { int num1 = (len1 >= 0)? input1[len1] : 0; int num2 = (len2 >= 0)? input2[len2] : 0; result[r] = (num1 + num2 + carry) % base; carry = (num1 + num2 + carry) / base; len1--; len2--; r++; } if(carry > 0) { result[r] = carry; r++; } result[r] = sign; reverse(result); return result; } /* * helper function for subtract * determine which number is larger of two positive numbers */ bool larger(int* input1, int* input2){ int len1 = bignum_length(input1); int len2 = bignum_length(input2); if (len1<=len2){ if (len1<len2){ //if input1 has less digit than input2 return false; } int i; for (i =0; i < len1; i++ ){//they have the same length if (input1[i]<input2[i]){ //if the same digit in input1 is smaller than that in input2 return false; } } } return true; //else input1 is indeed larger than/equal input2 (longer or every digit is no less than that in input2 } /* * helper function for subtract * subtract from the larger */ int* subtractLarger(int* input1, int* input2, int base){ //input1 is larger or equal than/to input2 and both positive int len1 = bignum_length(input1); int len2 = bignum_length(input2); int resultlength = ((len1 > len2) ? len1 : len2) + 2; int *result = (int *) malloc(sizeof(int) * resultlength); int r = 0; int carry = 0; int sign = -1; len1--; len2--; while(len1 >= 0 ) { int num1 = (len1 >= 0)? input1[len1]-carry : 0; int num2 = (len2 >= 0)? input2[len2] : 0; if (num1>=num2){ result[r] = (num1-num2); carry = 0; } else { result[r]= num1+base-num2; carry = 1; } len1--; len2--; r++; } if (result[r-1]==0){ result[r-1] = sign; } else { result[r] = sign; } reverse(result); return result; } /* * used to subtract two numbers with the same sign */ int* subtract (int* input1, int* input2, int base) { if (larger(input1,input2)){ return subtractLarger(input1, input2, base); } else { int* res = subtractLarger(input2, input1, base); //exchange input1 and input2, note the result is negative int sign = -2; //negative result res[bignum_length(res)] = sign; return res; } } /* * TODO * This function is where you will write the code that performs the heavy lifting, * actually performing the calculations on input1 and input2. * Return your result as an array of integers. * HINT: For better code structure, use helper functions. */ int* perform_math(int* input1, int* input2, char op, int base) { /* * this code initializes result to be large enough to hold all necessary digits. * if you don't use all of its digits, just put a -1 at the end of the number. * you may omit this result array and create your own. */ int resultlength = bignum_length(input1) + bignum_length(input2) + 2; int* result = (int*) malloc (sizeof(int) * resultlength); if(op == '+') { return add(input1, input2, base); } else if (op == '-'){ return subtract(input1, input2, base); } } /* * Print to "stderr" and exit program */ void print_usage(char* name) { fprintf(stderr, "----------------------------------------------------\n"); fprintf(stderr, "Usage: %s base input1 operation input2\n", name); fprintf(stderr, "base must be number between 2 and 36, inclusive\n"); fprintf(stderr, "input1 and input2 are arbitrary-length integers\n"); fprintf(stderr, "Two operations are allowed '+' and '-'\n"); fprintf(stderr, "----------------------------------------------------\n"); exit(1); } /* * MAIN: Run the program and tests your functions. * sample command: ./bignum 4 12 + 13 * Result: 31 */ int main(int argc, char** argv) { int input_base; int* input1; int* input2; int* result; if(argc != 5) { print_usage(argv[0]); } input_base = string_to_integer(argv[1]); if(!valid_base(input_base)) { fprintf(stderr, "Invalid base: %s\n", argv[1]); print_usage(argv[0]); } if(!valid_input(argv[2], input_base)) { fprintf(stderr, "Invalid input1: %s\n", argv[2]); print_usage(argv[0]); } if(!valid_input(argv[4], input_base)) { fprintf(stderr, "Invalid input2: %s\n", argv[4]); print_usage(argv[0]); } if(argv[3][0] != '-' && argv[3][0] != '+') { fprintf(stderr, "Invalid operation: %s\n", argv[3]); print_usage(argv[0]); } input1 = string_to_integer_array(argv[2]); input2 = string_to_integer_array(argv[4]); result = perform_math(input1, input2, argv[3][0], input_base); printf("Result: "); bignum_print(result); printf("\n"); exit(0); }，jqXHR.success()和jqXHR.error()回调方法被删除。

您可以改为使用jqXHR.complete()，jqXHR.done()和jqXHR.fail()。

这些方法最初作为选项回调添加到jQuery＆＃39; jqXHR.always()中，可以像这样使用

$.ajax

然而，由于用户之间存在一些混淆，他们后来还附带了具有相同名称的可链式方法

$.ajax({
    url      : 'mypage.php',
    success  : function() { ... },
    error    : function() { ... },
    complete : function() { ... }
});

这些方法自jQuery 1.8以来已被弃用，并且由于使用了Deferred对象而在jQuery 3.0中被完全删除，以及后来的承诺。

$.ajax().success( function() { ... }) .error( function() { ... }) .complete( function() { ... })，jqXHR.success()和jqXHR.error()被可链接的jqXHR.complete()，jqXHR.done()和jqXHR.fail()方法取代，选项回调现在还可以使用。

从jQuery 3.0开始，jQuery的延迟对象are also Promise/A+ compliant，这意味着它们是＆＃34;可以＆＃34;，并且可以与jqXHR.always()一起使用

then()

未捕获的TypeError：$ .ajax（...）。success不是函数

3 个答案: