Apex Code Development You need to sign in to do that
Don't have an account?
Raju I 5 System.NullPointerException: Attempt to de-reference a null object Error is in expression '{! search_now}' in component <apex:commandButton> in page accountsearch: External entry point
<apex:page Controller="c7">
<!---<apex:page contentType="c7">-->
<apex:form >
<apex:pageBlock title="My Search1">
<apex:inputText value="{!keyword}"/>
<apex:commandButton value="search" action="{! search_now}"/>
<apex:pageBlockTable value="{!results}" var="r">
<apex:column value="{!r.NAME}"/>
<apex:column value="{!r.TYPE}"/>
<apex:column value="{!r.PHONE}"/>
</apex:pageBlockTable>
</apex:pageBlock>
</apex:form>
</apex:page>
and apex
public class c7{
string keyword ;
List<account> results;
//public <account> results{get;set;}
public string getkeyword()
{
Return Keyword;
}
Public List<account> getresults()
{
return results;
}
//piblic void setKeyword (string input)
public void setkeyword (string input)
{
string keyword = input;
}
public pagereference search_now()
{
results=(list<account>)[FIND:keyword IN ALL FIELDS RETURNING account(NAME,PHONE)][0];
return null;
}
}
<!---<apex:page contentType="c7">-->
<apex:form >
<apex:pageBlock title="My Search1">
<apex:inputText value="{!keyword}"/>
<apex:commandButton value="search" action="{! search_now}"/>
<apex:pageBlockTable value="{!results}" var="r">
<apex:column value="{!r.NAME}"/>
<apex:column value="{!r.TYPE}"/>
<apex:column value="{!r.PHONE}"/>
</apex:pageBlockTable>
</apex:pageBlock>
</apex:form>
</apex:page>
and apex
public class c7{
string keyword ;
List<account> results;
//public <account> results{get;set;}
public string getkeyword()
{
Return Keyword;
}
Public List<account> getresults()
{
return results;
}
//piblic void setKeyword (string input)
public void setkeyword (string input)
{
string keyword = input;
}
public pagereference search_now()
{
results=(list<account>)[FIND:keyword IN ALL FIELDS RETURNING account(NAME,PHONE)][0];
return null;
}
}