Question

我不明白为什么call_script_id显示为字符串而不是变量中存储的内容。其他变量工作正常，但从字面上看，call_script_id设置id =“ call_script_id”，而call_reason_id设置正确的数字。

        $.ajax({
            url: "../selections/call-reasons.php",
            type: 'post',
            data: {company_uid:"<?php echo $row['company_uid']; ?>"},
            dataType: 'json',
            success:function(response){

                var len = response.length;

                $("#call_reasons").empty();

                for( var i = 0; i<len; i++){
                    var call_script_id = response[i]['call_script_id'];
                    var call_reason_id = response[i]['call_reason_id'];
                    var call_type = response[i]['call_type'];
                    var active = response[i]['active'];

                    $("#call_reasons").append("<tr><td href='../modals/call-types.php' class='call_reason_row' id=" + call_script_id + ">" + call_reason_id + "</td><td id=" + call_script_id + "></td><td id=" + call_script_id + "></td><td><i class='far fa-edit'></i><i class='far fa-calendar-alt'></i><i class='far fa-trash-alt call_reason_trash' id=" + call_reason_id + "></i></td></tr>");
                }
                // Brings up the pop up to edit call reasons/types
                $(".call_reason_row").click(function() {
                    $('#main-content',parent.document).load($(this).attr('href'), {call_reason_id: this.id, company_uid: "<?php echo $row["company_uid"];?>", active: active});
                });
            },
            error: function (xhr, status, error) {
                alert(xhr.responseText);
            }
        });

        $sql = "SELECT cs.id, cs.call_reason, sct.call_type FROM call_script AS cs INNER JOIN selection_call_types AS sct ON cs.call_reason = sct.id WHERE cs.company_uid = '$company_uid'";

        $result = mysqli_query($conn, $sql);

        $my_array = array();

        while($row = mysqli_fetch_array($result) ){
            $call_script_id = $row['id'];
            $call_reason_id = $row['call_reason'];
            $call_type = $row['call_type'];
            $active = $row['active'];

            $my_array[] = array("call_script_id" => call_script_id, "call_reason_id" => $call_reason_id, "call_type_id" => $call_type, "active" => $active);
        }

Answer 1

我猜我希望问题出在ajax中，因为我不太熟悉该语言。相反，这是因为我在最后一行的$ my_array中缺少变量call_reason_id上的美元符号。

为什么变量显示为字符串？

1 个答案: