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Can a string with the word NULL in it ever be interpreted as a NULL value?

由 て烟熏妆下的殇ゞ 提交于 2021-02-19 03:44:09
问题 Looking for explanations for an error in a db insert, or update, I was told the problem is a customer name like: "BOB NULL-Creations" . What I have read about NULL doesn't allow me to accept that explanation so easily. However I have been wrong about other things before. Is it ever possible for the string 'NULL' to ever be interpreted as a NULL value in a situation where it is part of a string? A little more information The data is being ultimately viewed on an iPad so I am told the NULL

What if a null character is present in the middle of a string?

由 烈酒焚心 提交于 2021-02-18 16:57:46
问题 I understand that the end of a string is indicated by a null character, but i cannot understand the output of the following code. #include <stdio.h> #include <string.h> int main(void) { char s[] = "Hello\0Hi"; printf("%d %d", strlen(s), sizeof(s)); } OUTPUT: 5 9 If strlen() detects the end of the string at the end of o, then why doesn't sizeof() do the same thing? Even if it doesn't do the same thing, isn't '\0' A null character (i.e, only one character), so shouldn't the answer be 8? 回答1:

Error: invalid input syntax for integer: “”

由 ￡可爱￡侵袭症＋ 提交于 2021-02-18 11:12:14
问题 I have this table tbl_buku : id_buku judul_buku tahun_buku 1 Bioogi 2010 2 Fisika 2010 3 Informatika 2012 4 Kimia 2012 I use query like this, but I am getting an error: select case when t1.tahun_buku=t2.tahun_buku then '' else t1.tahun_buku end tahun_buku,t1.judul_buku from tbl_buku t1 left join tbl_buku t2 on t1.id_buku-1=t2.id_buku; I want to show table like this: tahun_buku judul_buku 2010 Biologi Fisika 2012 Informatika Kimia How to achieve this? 回答1: I think the problem in your query is

Error: invalid input syntax for integer: “”

由 為{幸葍}努か 提交于 2021-02-18 11:12:07
问题 I have this table tbl_buku : id_buku judul_buku tahun_buku 1 Bioogi 2010 2 Fisika 2010 3 Informatika 2012 4 Kimia 2012 I use query like this, but I am getting an error: select case when t1.tahun_buku=t2.tahun_buku then '' else t1.tahun_buku end tahun_buku,t1.judul_buku from tbl_buku t1 left join tbl_buku t2 on t1.id_buku-1=t2.id_buku; I want to show table like this: tahun_buku judul_buku 2010 Biologi Fisika 2012 Informatika Kimia How to achieve this? 回答1: I think the problem in your query is

Django: Want to display an empty field as blank rather displaying None

由 一笑奈何 提交于 2021-02-18 08:53:46
问题 I have a template called client_details.html that displays user , note and datetime . Now sometimes, a client may not have an entry for user, note and datetime. What my program will do instead is display None if these fields are empty. I do not want the to display None. If a field has no value I don't want to see any value e.g. let it be blank if possible instead of displaying None. views.py @login_required def get_client(request, client_id = 0): client = None try: client = models.Client

char * return (null) from a function in C [duplicate]

由 こ雲淡風輕ζ 提交于 2021-02-17 07:11:16
问题 This question already has answers here : Returning an array using C (8 answers) Closed last month . I was trying to append two const char* from a function but it returns (null) . What should I do now? I'm working in VS Code on Ubuntu 20.04 using GCC Compiler v9.3.0. Code #include <string.h> #include <stdio.h> char *JoinChars(const char *a, const char *b) { char buffer[strlen(a) + strlen(b) + 1]; strcpy(buffer, a); strcat(buffer, b); return buffer; } int main() { const char *b1 = "Hello\t";

char * return (null) from a function in C [duplicate]

由 与世无争的帅哥 提交于 2021-02-17 07:10:23
问题 This question already has answers here : Returning an array using C (8 answers) Closed last month . I was trying to append two const char* from a function but it returns (null) . What should I do now? I'm working in VS Code on Ubuntu 20.04 using GCC Compiler v9.3.0. Code #include <string.h> #include <stdio.h> char *JoinChars(const char *a, const char *b) { char buffer[strlen(a) + strlen(b) + 1]; strcpy(buffer, a); strcat(buffer, b); return buffer; } int main() { const char *b1 = "Hello\t";

android PopupWindow

由 这一生的挚爱 提交于 2021-02-13 05:30:48
1 PopupWindow window = new PopupWindow(activity);//的到对象 2 window.setBackgroundDrawable(activity.getResources().getDrawable(R.drawable.configlistviewbg));//设置背景 //设置PopupWindow显示和隐藏时的动画 window.setAnimationStyle(R.style.AnimationFade); View view=LayoutInflater.from(activity).inflate(joker.housemananger.activity.R.layout.addhouseactivity_popuwindow, null,true);//将popu的布局文件实例出 long wid=activity.getWindowManager().getDefaultDisplay().getWidth(); //的到屏幕的宽 long hd=activity.getWindowManager().getDefaultDisplay().getHeight();//得到屏幕的高 View call=view.findViewById(R.id.addhouse_popuwindow_bt_callphone);/

MApping JSON to array return null

由 本秂侑毒 提交于 2021-02-11 13:22:09
问题 When I use this line I get : AudDto[] auditDtos = objectMapper.readValue(responseBodyAsString, AudDto[].class); I got : com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of `` out of START_OBJECT token So I add this line : objectMapper.enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY); After that I got : com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field “errMsg” So I used : objectMapper.configure

MApping JSON to array return null

由 帅比萌擦擦* 提交于 2021-02-11 13:21:57
问题 When I use this line I get : AudDto[] auditDtos = objectMapper.readValue(responseBodyAsString, AudDto[].class); I got : com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of `` out of START_OBJECT token So I add this line : objectMapper.enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY); After that I got : com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field “errMsg” So I used : objectMapper.configure
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