Can not convert 'string' to 'int'

Can not convert 'string' to 'int'

By : Ivo Carvalho
Date : October 28 2020, 08:00 PM
this one helps. At first two important hints:
Your code is open to SQL injection attacks. This is dangerous. Please use parameterized queries. Don't use SELECT * but state explicitly which columns to return. This way you avoid unnecessary traffic and can be sure to get the columns in the order you expect.
code :
textBoxFileName.Text = mdr.GetString(mdr.GetOrdinal("FileName"));

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How can I convert a query string into a a shorter alphanumeric string (and convert it back again) in PHP?

How can I convert a query string into a a shorter alphanumeric string (and convert it back again) in PHP?

By : Katsuma Narisawa
Date : March 29 2020, 07:55 AM
wish helps you What you are seeking is not a hash - since hash is a one-way function in common case. Here is a possible solution - use both base64 encryption and something like parameters map, so you will able to get shorter file name because you'll not store parameters names, only values:
code :
class Holder
   const NAME_PARAM_DELIMIER = '|';

   public static function getParametersMap()
      return [
        0 => 'count',
        1 => 'deshortifyurl',
        2 => 'extracturl',
        3 => 'filterhashtagsend',
        4 => 'filterscreennames',
        5 => 'filterspecifiedhashtag',
        6 => 'filterurls',
        7 => 'hashtag',
        8 => 'inputType',

   public static function getParamsByName($sName, $bReturnAsArray=true)
      $rgParams = @array_combine(self::getParametersMap(), explode(self::NAME_PARAM_DELIMIER, base64_decode($sName)));
         return null;
      return $bReturnAsArray?$rgParams:http_build_query($rgParams);

   public static function getNameByParams($sQuery)
      parse_str($sQuery, $rgParams);
      return base64_encode(join(self::NAME_PARAM_DELIMIER, array_values($rgParams)));
$sQuery = 'inputType=timeline&count=50&hashtag=%23li&filterspecifiedhashtag=1&filterhashtagsend=1&filterscreennames=1&extracturl=1&deshortifyurl=1&filterurls=1';

$sName  = Holder::getNameByParams($sQuery);
$rgData = Holder::getParamsByName($sName);
var_dump($sName); //NTB8MXwxfDF8MXwxfDF8I2xpfHRpbWVsaW5l
convert a string to a number, do math on it, then convert it back and give the span that value as a string

convert a string to a number, do math on it, then convert it back and give the span that value as a string

By : Szekely Istvan
Date : March 29 2020, 07:55 AM
Does that help i am trying to take entered values from textboxes and convert them to strings, total the amount, and then set the innerHTML of a to that value. , You're Doing It Wrong
For starters.
code :
function addBills(){
        hundreds = parseInt($("#input100").innerHTML(),10)*100;
function addBills(){
   var hundreds = Number(document.getElementById('input100').value)*100
   ,   fifties = Number(document.getElementById('input50').value)*50
   ,   twenties = Number(document.getElementById('input20').value)*20
   ,   tens = Number(document.getElementById('input10').value)*10
   ,   fives = Number(document.getElementById('input5').value)*5
   ,   ones = Number(document.getElementById('input1').value)
   ,   regTotal = hundreds+fifties+twenties+tens+fives+ones
   ; //end local
Can't convert string to an integer in VBA or by Excel formulas (failure to convert or wrong number even if string is all

Can't convert string to an integer in VBA or by Excel formulas (failure to convert or wrong number even if string is all

By : Amorino89
Date : March 29 2020, 07:55 AM
To fix the issue you can do You can identify this error (overflow) in your error-handling block using the Description property of the intrinsic Err object:
code :
MsgBox "Failed to convert """ & v1 & """ to an integer." & vbCrLf & Err.Description, , "Aborting - Failed Conversion" 
Dim tempArray As Variant
Dim arraySize As Double
arraySize = (Len(v1) - 1)
ReDim tempArray(arraySize)
Sub foo()
Dim lrg As New cLarge

MsgBox lrg.LargeAdd("105501008962100001", "205501231962100003")
End Sub
'### Class module for adding very large (> Decimal precision) values
'    http://tushar-mehta.com/misc_tutorials/project_euler/LargeNumberArithmetic.htm
'    Modified by David Zemens, 9 December 2015
Option Explicit

Public cDecMax As Variant, cDecMaxLen As Integer, cSqrDecMaxLen As Integer
Private pVal As String

Public Sub Class_Initialize()
    Static Initialized As Boolean
    If Initialized Then Exit Sub
    Initialized = True
    cDecMax = _
        CDec(Replace("79,228,162,514,264,337,593,543,950,335", ",", ""))
            'this is 2^96-1, the limit on Decimal precision data in Excel/VBA
    cDecMaxLen = Len(cDecMax) - 1
    cSqrDecMaxLen = cDecMaxLen \ 2
End Sub

Function Ceil(X As Single) As Long
    If X < 0 Then Ceil = Fix(X) Else Ceil = -Int(-X)
End Function

Private Function addByParts(ByVal Nbr1 As String, ByVal Nbr2 As String) _
    As String
    Dim NbrChunks As Integer
    If Len(Nbr1) > Len(Nbr2) Then _
        Nbr2 = String(Len(Nbr1) - Len(Nbr2), "0") & Nbr2 _
    Else _
        Nbr1 = String(Len(Nbr2) - Len(Nbr1), "0") & Nbr1
    NbrChunks = Ceil(Len(Nbr1) / cDecMaxLen)
    Dim I As Integer, OverflowDigit As String, Rslt As String
    OverflowDigit = "0"
    For I = NbrChunks - 1 To 0 Step -1
        Dim Nbr1Part As String
        Nbr1Part = Mid(Nbr1, I * cDecMaxLen + 1, cDecMaxLen)
        Rslt = CStr(CDec(Nbr1Part) _
            + CDec(Mid(Nbr2, I * cDecMaxLen + 1, cDecMaxLen)) _
            + CDec(OverflowDigit))
        If Len(Rslt) < Len(Nbr1Part) Then
            Rslt = String(Len(Nbr1Part) - Len(Rslt), "0") & Rslt
            OverflowDigit = "0"
        ElseIf I = 0 Then
        ElseIf Len(Rslt) > Len(Nbr1Part) Then
            OverflowDigit = Left(Rslt, 1): Rslt = Right(Rslt, Len(Rslt) - 1)
            OverflowDigit = "0"
            End If
        addByParts = Rslt & addByParts
        Next I
End Function

Function LargeAdd(ByVal Nbr1 As String, ByVal Nbr2 As String) As String
    If Len(Nbr1) <= cDecMaxLen And Len(Nbr2) <= cDecMaxLen Then
        LargeAdd = CStr(CDec(Nbr1) + CDec(Nbr2))
        Exit Function
        End If
    If Len(Nbr1) > cDecMaxLen Then LargeAdd = addByParts(Nbr1, Nbr2) _
    Else LargeAdd = addByParts(Nbr2, Nbr1)
End Function
How to convert a time whch given in string and find the difference in java and convert back to string

How to convert a time whch given in string and find the difference in java and convert back to string

By : Feudimonster
Date : March 29 2020, 07:55 AM
I hope this helps you . You can use the below method to get the time difference, just pass HH:MM without the timezone.
code :
public long getTimeDifference(String time1, String time2) throws ParseException{
        SimpleDateFormat format = new SimpleDateFormat("HH:mm");
        Date date1 = format.parse(time1);
        Date date2 = format.parse(time2);
        long difference = date2.getTime() - date1.getTime();
        return difference/1000;
Cannot directly convert number to hex null-terminated string, has to convert to std::string then use .c_str()

Cannot directly convert number to hex null-terminated string, has to convert to std::string then use .c_str()

By : user2908997
Date : March 29 2020, 07:55 AM
this one helps. Your main problem is that you define a variable on the stack, locally in the function, and then return it.
After the function returns, the char* will point to "somewhere", to an undefined position. That is a major bug. You have also other bugs that have been commented on already. Like sprintf(szHex, "0x%s", szHex);, which is undefined behaviour (UB) or sprintf(szHex, "%s%s", hex[nTemp], szHex); which has the same problem + additionally a wrong format string.
code :
#include <iostream>
#include <string>
#include <iomanip>
#include <sstream>

std::string toHexString(unsigned int hexValue)
    std::ostringstream oss;
    oss << "0x" << std::hex << hexValue;
    return std::string(oss.str());

int main()
    std::cout << toHexString(15) << '\n';
    // or directly
    std::cout << "0x" << std::hex << 15 << '\n';

    return 0;
#include <stdio.h>
#include <iostream>

char* toHexCharP(unsigned int hexValue, char *outBuffer, const size_t maxSizeOutBuffer)
    return outBuffer;    

constexpr size_t MaxBufSize = 100U;

int main()
    char buf[MaxBufSize];
    std::cout << toHexCharP(15, buf, MaxBufSize) << '\n';
    return 0;
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