warnings

问题 The following code is throwing the mysterious Warnings. I can't understand what they mean. What do these errors indicate and how to eradicate them? require "conn.php"; $q = mysqli_stmt_init($dbconn); $query = "SELECT users.userid FROM users WHERE users.email = ? "; mysqli_stmt_prepare($q, $query); mysqli_stmt_bind_param($q, "s", $email); mysqli_stmt_execute($q); $result = mysqli_stmt_get_result($q); if (mysqli_num_rows($result) == 0) { $q = mysqli_stmt_init($dbconn); $query = "INSERT INTO

问题 The following code is throwing the mysterious Warnings. I can't understand what they mean. What do these errors indicate and how to eradicate them? require "conn.php"; $q = mysqli_stmt_init($dbconn); $query = "SELECT users.userid FROM users WHERE users.email = ? "; mysqli_stmt_prepare($q, $query); mysqli_stmt_bind_param($q, "s", $email); mysqli_stmt_execute($q); $result = mysqli_stmt_get_result($q); if (mysqli_num_rows($result) == 0) { $q = mysqli_stmt_init($dbconn); $query = "INSERT INTO

问题 I have a xlsx file with a column containing Dates in the format: "01.01.1900 09:01:25". The file is password protected so I convert it to a dataframe by means of win32com.client library. Here is the code: import pandas as pd import win32com.client xlApp = win32com.client.Dispatch("Excel.Application") xlApp.DisplayAlerts = False xlwb = xlApp.Workbooks.Open(File, False, True, None, " ") #Open Workbook password " " xlws = xlwb.Sheets("Sheet 1") #Open Sheet 1 #Get table dimensions LastRow = xlws

问题 my app was killed,and the log :libpng warning: Ignoring bad adaptive filter type. I want to know the reason,and the solution. 回答1: The question gives insufficient information to provide a complete answer. The "bad adaptive filter type" message means that the PNG file is either corrupted or was generated by a buggy encoder. Each scanline of a PNG file begins with a "filter type" byte which must be 0, 1, 2, 3, or 4. 回答2: I had this issue before, the image was corrupted by unknown reason, the

问题 I'm running a PHP script and continue to receive errors like: Notice: Undefined variable: my_variable_name in C:\wamp\www\mypath\index.php on line 10 Notice: Undefined index: my_index C:\wamp\www\mypath\index.php on line 11 Line 10 and 11 looks like this: echo "My variable value is: " . $my_variable_name; echo "My index value is: " . $my_array["my_index"]; What is the meaning of these error messages? Why do they appear all of a sudden? I used to use this script for years and I've never had

问题 I'm refactoring a library and trying to get rid of many gcc warnings. The big part of these warning are about signed / unsigned comparison and are related to the usage of size_t . The library works on 64 bit Linux systems. A programmer used -1 as the special value similar to std::string::npos . There are many places in the library where code looks like this: class AnnotationBase { public: size_t m_offset = -1; size_t m_length = -1; } ... AnnotationBase foo(const std::string& text, const

问题 I'm refactoring a library and trying to get rid of many gcc warnings. The big part of these warning are about signed / unsigned comparison and are related to the usage of size_t . The library works on 64 bit Linux systems. A programmer used -1 as the special value similar to std::string::npos . There are many places in the library where code looks like this: class AnnotationBase { public: size_t m_offset = -1; size_t m_length = -1; } ... AnnotationBase foo(const std::string& text, const

问题 In Struts 2 application we use log4j for logging. I want to log only warn but when i try to use in my log4j.properties log4j.rootLogger=warn, stdout it prints error and fatal log too. I want only warn log. I read the levels of log in some tutorials. So I know why error and fatal prints, because they are less priority than warn. How to print warn log only in my console? My Java code: import org.apache.log4j.Logger; public class LogClass { private static final org.apache.log4j.Logger log =

问题 In Struts 2 application we use log4j for logging. I want to log only warn but when i try to use in my log4j.properties log4j.rootLogger=warn, stdout it prints error and fatal log too. I want only warn log. I read the levels of log in some tutorials. So I know why error and fatal prints, because they are less priority than warn. How to print warn log only in my console? My Java code: import org.apache.log4j.Logger; public class LogClass { private static final org.apache.log4j.Logger log =

问题 I have a large block of code that is, at some point somewhere, generating a setting with copy warning in pandas (this problem). I know how to fix the problem, but I can't find what line number it is! Is there a way to back out the line number (apart from brute force methods like debug-stepping or putting in multiple prints)? The only output I get is the below, which doesn't go up the stack to my code: C:\Anaconda3\lib\site-packages\pandas\core\frame.py:2302: SettingWithCopyWarning: A value is