Problem with delete statement

Hi there, I have created this code below.  The select statement works, however the delete statement does not work.
DECLARE
@CategoryASVARCHAR(255)
DECLARE
@NameASVARCHAR(255)
DECLARE
@Name1ASVARCHAR(255)
DECLARE
@ParentTypeAsint
SET
@Category='General
Building Data'
SET
@Name='Purchase
Date'
SET
@Name1='Purchase
Cost'
Set
@ParentType=20
Select
*fromtbAttributeValueA
Join
tbAttributeTemplateDefinitionLinkATDLONATDL.AttributeTemplateDefinitionLinkID=A.AttributeTemplateDefinitionLinkID
Join
tbAttributeTemplateDefinitionATDONATD.AttributeTemplateDefinitionID=ATDL.AttributeTemplateDefinitionID
JOIN
tbAttributeSetSONS.AttributeSetID=ATDL.LinkedParentID
Where
S.ParentType=@ParentType
AND
S.Name=@Category
AND
([email protected]=@Name1)
AND
A.Value=0
delete
fromtbAttributeValueA
Join
tbAttributeTemplateDefinitionLinkATDLONATDL.AttributeTemplateDefinitionLinkID=AttributeTemplateDefinitionLinkID
Join
tbAttributeTemplateDefinitionATDONATD.AttributeTemplateDefinitionID=ATDL.AttributeTemplateDefinitionID
JOIN
tbAttributeSetSONS.AttributeSetID=ATDL.LinkedParentID
Where
S.ParentType=@ParentType
AND
S.Name=@Category
AND
([email protected]=@Name1)
AND
A.Value=0
The delete statement fails on the first join.     Incorrect Syntax near 'A'.  What am I doing wrong?  As I have only changed the select * to delete.  So therefore does this mean I need to put the delete statement into a
sql sub query? Or is there another way of doing this?

Try this:
DELETE FROM tbAttributeValue
FROM tbAttributeTemplateDefinitionLink ATDL
JOIN tbAttributeValue t
ON ATDL.AttributeTemplateDefinitionLinkID = t. AttributeTemplateDefinitionLinkID
JOIN tbAttributeTemplateDefinition ATD
ON ATD.AttributeTemplateDefinitionID = ATDL.AttributeTemplateDefinitionID
JOIN tbAttributeSet S
ON S.AttributeSetID = ATDL.LinkedParentID
WHERE S.ParentType = @ParentType
AND S.NAME = @Category
AND ( ATD.NAME = @Name
OR ATD.NAME = @Name1 )
AND A.Value = 0

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        System.out.println( "Please write a sentence " );
        line = scan.nextLine();
        total=line.length(); //Gesamtanzahl an Zeichen des Satzes
        for (int counter=0; counter<total; counter++)
          letter = line.charAt(counter); //ermitteln des Buchstabens an einer bestimmten Position des Satzes
          switch (letter)
            case 'A': case 'a':
            case 'E': case 'e':
            case 'I': case 'i':
            case 'O': case 'o':
            case 'U': case 'u':
              countV++;
              break;
            case 'B': case 'b': case 'C': case 'c': case 'D': case 'd': case 'F': case 'f': case 'G': case 'g': case 'H': case 'h':
            case 'J': case 'j': case 'K': case 'k': case 'L': case 'l': case 'M': case 'm': case 'N': case 'n': case 'P': case 'p':
            case 'Q': case 'q': case 'R': case 'r': case 'S': case 's': case 'T': case 't': case 'V': case 'v': case 'W': case 'w':
            case 'X': case 'x': case 'Y': case 'y': case 'Z': case 'z':
              countC++;
              break;
            case ' ':
              countS++;
              break;
            case ',': case '.': case ':': case '!': case '?':
              countP++;
              break;
            case 'Ä': case 'ä': case 'Ö': case 'ö': case 'Ü': case 'ü':
              countU++;
              break;
        System.out.println( "Total amount of characters:\t" + total );
        System.out.println( "Number of consonants:\t\t" + countC );
        System.out.println( "Number of vocals:\t\t" + countV );
        System.out.println( "Number of umlauts:\t\t" + countU );
        System.out.println( "Number of spaces:\t\t" + countS );
        System.out.println( "Number of punctuation chars:\t" + countP );
    }

    WRE wrote:
    •In general I’m not very happy with this huge list of “cases”. How would you solve a problem like this? Is there a more convenient/elegant way?I've been doing this a lot lately myself evaluating documents with 20 or so million words. Few tips:
    1. Regular expressions can vastly reduce the list of cases. For example you can capture all letters from a to z or A to Z as follows [a-zA-Z]. To match a single character in a String you can then make use of the Pattern and Matcher classes, and incorporate the regular expression. e.g.
      //Un-compiled code, may contain errors.
      private Pattern letterPattern = Pattern.compile("[a-zA-Z]");
      public int countNumberOfLettersInString(final String string) {
        int count = 0;
        Matcher letterMatcher = letterPattern.matcher(string);
        while(letterMatcher.find()) {
          count++;
        return count;
      }2. As mentioned above, Sets are an excellent choice. Simply declare a static variable and instantiate it using a static initializer block. Then loop over the String to determine if the character is in the given set. e.g.
      //Un-compiled code, may contain errors.
      private static Set<Character> macrons = new HashSet<Character>();
      static {
        macrons.add('ä');
        macrons.add('ö');
        macrons.add('ü');
      public int countNumberOfMacronsInString(final String string) {
        int count = 0;
        for(char c : string.toCharArray()) {
          if(macrons.contains(c) {
            count++;
        return count;
      }Mel

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