You can’t use not like directly in a show tables statement, but you could put it in a where clause: SHOW TABLES WHERE tables_in_db NOT LIKE ‘wp_2_%’; solved NOT LIKE not working with SHOW [duplicate]
Taking the initial code in the link, and adding parentheses helps to break down the If and Else structure: @Echo Off For %%Z In (“%ATTR%”) Do ( If “%%~aZ” GEq “d” ( Echo Directory ) Else ( If “%%~aZ” GEq “-” ( Echo File ) Else ( Echo Inaccessible ) ) ) Pause So to … Read more
C++ error: Reference to non-static member function must be called, Please help, don’t know whats literally wrong [closed] solved C++ error: Reference to non-static member function must be called, Please help, don’t know whats literally wrong [closed]
You are appending mainmenu.tbxSelected whereas I suspect that your intention was to append mainmenu.tbxSelected.Text The former would probably result in a fully qualified typename in your WHERE clause, which would contain commas. As an aside, you should be aware that constructing SQL in this way leaves you potentially open to SQL Injection attacks. If that’s … Read more
I assume you want something like: def answer(plaintext): words = {“a”:’100000′,”b”:’110000′,”c”:’100100′,”d”:’100110′,”e”:’100010′,”f”:’110100′,”g”:’110110′,”h”:’110010′,”i”:’010100′,”j”:’010110′,”k”:’101000′,”l”:’111000′,”m”:’101100′,”n”:’101110′,”o”:’101010′,”p”:’111100′,”q”:’111110′,”r”:’111010′,”s”:’011100′,”t”:’011110′,”u”:’101001′,”v”:’111001′,”w”:’010111′,”x”:’010111′,”y”:’101011′,”z”:’101011′} inputList = plaintext.split(‘,’) for word in inputList: print words[word] text = “j,o,s,e” answer(text) You had a bunch of typos in your dict, pay attention to the traceback, it tells you exactly what’s wrong. You also never actually called the function you defined. You probably … Read more
$ python test.py File “test.py”, line 11 While not x==0: ^ SyntaxError: invalid syntax This is a bug in Python. The caret should be pointing to the capital ‘W’ in ‘While’. You must spell ‘while’ with all lowercase letters. You have some other typos as well: File “test.py”, line 17 for divisor in the range … Read more
First of all, paste your code correctly… You are missing </table> and </form> in the end… Second of all, you were missing a simple apostrophe and a semi-colon in the line $result = mysquli_query($con, ‘select * from account where username=”‘.$username.'” and password=”‘.$password.'”) Also, you should use mysqli_query instead of mysqlui_query… Typo! Your code should be … Read more
It appears that your issue is happening because you have a space between “!” and “=” on all of the lines that you tagged. You’re attempting to use the inequality operator, which is “!=”, or an exclamation mark followed by an equals sign with no space. When you have the exclamation mark by itself, it … Read more
Try this: </div> <input className=”channel-search+__input__text” placeholder=”Search perhaps?” type=”text” value={query} onChange={onSearch} /> </div> 1 solved Unexpected token, expected “…” in ChannelSearch.jsx [closed]
If the code is a part of bbpress I think the following link could prove to be helpful. http://codex.bbpress.org/bbp_list_forums/ 0 solved syntax error, unexpected ‘false’ (T_STRING)? [closed]
Here is the error You are saying var second = document.getElemntById(‘cpassword’).value; it should be var second = document.getElementById(‘cpassword’).value; 0 solved Error with document.getElementById: Uncaught TypeError: undefined is not a function [closed]
You’re missing a while after the do block. They go together. The syntax for do … while in Java is: do { // some statements } while (/*condition*/); More info 3 solved Online prompt Java : external error or my own error?
Your minSdkVersion is 10. appcompat-v7 version 26.0.0-alpha1 no longer supports back that far; it only supports to API Level 14. Either: Raise your app’s minSdkVersion to 14, or Stick with an older version of appcompat-v7, one that still supports a minSdkVersion of 10 or lower 4 solved Execution failing during :app:processDebugManifest gradle task in Android … Read more
You simply cannot define a function inside another function. You are attempting to define echoer inside main. Instead, it should go outside. You should also know that ‘ is used for character literals and ” is used for string literals. Since ‘) ‘ has more than one character in it, it needs to be a … Read more
parseInt(number, 10)= b; So you are saying to assigning b to parseInt. You have it reversed. var b = parseInt(number, 10) You have the problem in multiple places. solved I get an invalid left hand side in assignment with my code [closed]