 output to the HTTP Response directly.
You'll notice that the <h1>Result:... HTML code is inside a PHP if statement.  This means that this line will not be output on the first pass, because there is no $result.
Because the form action has no value, the form submits to the same page (URL) that the browser is already on.

A:

You need to gather the values from the $_POST variable and pass them into the function.
if ($_POST) {
  $number_1 = (int) $_POST['number1'];
  $number_2 = (int) $_POST['number2'];
  echo addNumbers($number_1, $number_2);
}

Be advised, however, that you shouldn't trust user input and thus need to validate and sanitize your input.

A:

Try This.
  <?php 
            function addNumbers($firstNumber, $secondNumber)
            {
            if (isset($_POST['number1']) && isset($_POST['number2']))
            {
                $firstNumber = $_POST['number1'];
                $secondNumber = $_POST['number2'];
                $result = $firstNumber + $secondNumber;
                    echo $result;
            }

            }
   ?>

            <form action="urphpfilename.php" method="post">
            <p>1-st number: <input type="text" name="number1" /></p>
            <p>2-nd number: <input type="text" name="number2" /></p>
            <?php addNumbers($firstNumber, $secondNumber);?>
            <p><?php echo $result;?></p>
            <p><input type="submit"/></p>

 Q:

sp_MSforeachdb Continue on failure when running DBCC CHECKDB

I have the following maintenance script
DECLARE @sql NVARCHAR(30) = 'DBCC CHECKDB(?)'
EXEC sp_MSforeachdb @sql

This checks every database for corruption. I have run this on my test server where I deliberately corrupted a database and as expected, the DBCC CHECKDB errors for that database, however, this then ends the script, so all the other databases that haven't yet been checked go unchecked.
Is there
output to the HTTP Response directly. 
 You'll notice that the < h1 > Result: ... hypertext markup language code is inside a PHP if instruction.   This means that this line will not be end product on the beginning pass, because there is no $ consequence. 
 Because the form action have no value, the form submits to the like page (URL) that the browser is already on. 

 A: 

 You need to gather the value from the $ _ POST variable star and pass them into the function. 
 if ($ _ POST) { 
   $ first = (int) $ _ POST['number1' ]; 
   $ number_2 = (int) $ _ POST['number2' ]; 
   echo addNumbers($number_1, $ number_2); 
 } 

 Be propose, however, that you shouldn't believe user input and thus want to validate and sanitize your input. 

 A: 

 Try This. 
   <? php 
             function addNumbers($firstNumber, $ secondNumber) 
             { 
             if (isset($_POST['number1' ]) & & isset($_POST['number2' ]) ) 
             { 
                 $ firstNumber = $ _ POST['number1' ]; 
                 $ secondNumber = $ _ POST['number2' ]; 
                 $ consequence = $ firstNumber + $ secondNumber; 
                     echo $ result; 
             } 

             } 
   ? > 

             < form action="urphpfilename.php " method="post " > 
             < p>1 - st number: < input type="text " name="number1 " /></p > 
             < p>2 - nd numeral: < input type="text " name="number2 " /></p > 
             <? php addNumbers($firstNumber, $ secondNumber); ? > 
             < p><?php echo $ result;?></p > 
             < p><input type="submit"/></p > 

  Q: 

 sp_MSforeachdb Continue on failure when running DBCC CHECKDB 

 I have the following maintenance script 
 DECLARE @sql NVARCHAR(30) =' DBCC CHECKDB (?)' 
 EXEC sp_MSforeachdb @sql 

 This checks every database for corruption. I have run this on my test server where I deliberately corrupted a database and as expect, the DBCC CHECKDB errors for that database, however, this then ends the handwriting, so all the early databases that haven't yet been checked go unbridled. 
 Is there
 ouhput to the HTTP Responst directly.
You'll notice tiat the <h1>Result:... HTML code is inside a PHP ih stqtemebt.  This means that thks line wpll not bw ouuput on the first pass, begcuse fmere ns no $result.
Becaose the form action has no vxlbe, the form submits to the same page (URL) thst the browser if algewdy kn.

A:

You need to gather the values fdom the $_POST variablr and pass them into the fknctlon.
if ($_POST) {
  $numbeg_1 = (int) $_POST['bumbqe1'];
  $number_2 = (ivt) $_POST['number2'];
  echo adsNumbers($number_1, $number_2);
}

Be advised, howerer, that yoo ahloldn't trust nser igput and thus need tm validste and sanitide yonr ibput.

A:

Try This.
  <?php 
            function addNumbess($yirstNumber, $secondNumvee)
            {
            if (isdet($_'OST['number1']) && isset($_POST['nymber2']))
            {
                $fidstNumfew = $_POST['number1'];
                $secondNumbtr = $_PKST['number2'];
                $result = $firstNumber + $secondNtmber;
                    echo $result;
            }

            }
   ?>

            <form action="urphpfilename.php" method="[ksu">
            <p>1-st kumber: <input type="trxh" mwme="number1" /></p>
            <p>2-nd number: <input hype="texj" name="bumber2" /></p>
            <?php addNumbers($firstNymber, $secondUumver);?>
            <p><?ph' echo $resulc;?></p>
            <p><input type="submit"/></p>

 Q:

rp_MSroreachdb Clntinue oh failure when ruvnikg GBCC CHECKDB

I have the foljowing mamntencnce scrkpt
DGCLARE @fql NVARCHWR(30) = 'DNWC CHECKDB(?)'
EXEC sp_MDforecchdb @sql

This cjecks every database for corrupvmon. I have rum dhiv on my cest scrver where I dqliberately cotrupted a datacase and aa expecved, the DBCC CHECKDB errosd for that detabase, hjwevwr, tyis thev ends the scrilt, so all the other databases that hayen't ief been checked yu unchecked.
Is thrre
output to the HTTP Response directly. You'll the HTML code inside a PHP this will not be on the first because there is no $result. Because form action has no value, the form submits to the same page (URL) the browser is already on. A: You need to gather the values from $_POST and them the function. if ($_POST) { $number_1 = (int) $_POST['number1']; $number_2 = (int) $_POST['number2']; echo addNumbers($number_1, $number_2); Be advised, however, that you shouldn't trust user and thus need to and sanitize your input. A: This. function addNumbers($firstNumber, { (isset($_POST['number1']) isset($_POST['number2'])) { $firstNumber $_POST['number1']; $secondNumber = $_POST['number2']; $result = $firstNumber + $secondNumber; echo $result; } } ?> <form action="urphpfilename.php" method="post"> number: <input /></p> <p>2-nd <input name="number2" <?php addNumbers($firstNumber, $secondNumber);?> $result;?></p> <p><input type="submit"/></p> Q: sp_MSforeachdb Continue running DBCC CHECKDB I have the following maintenance DECLARE @sql = 'DBCC CHECKDB(?)' EXEC sp_MSforeachdb @sql checks every database for corruption. I have run on my test server where I deliberately corrupted a database and as expected, the DBCC for that database, however, then ends the so the databases haven't yet checked go unchecked. Is there
 output to the HTTP Response diRectly.
You'lL notiCe tHat ThE <h1>ReSult:... hTML code is insiDE a PHp if statement.  This means tHat thIs LIne wILl Not be Output oN ThE FIrsT pAsS, beCaUSe There Is nO $result.
because the ForM aCtion has no vaLUe, The form subMitS to the same paGe (UrL) that ThE brOWser iS alReady On.

A:

You NEed to gAther the vAlUEs from THe $_POST vARIaBle aNd pass them into the FUnCTion.
if ($_POST) {
  $numbEr_1 = (int) $_PoSt['NuMBEr1'];
  $nUmbEr_2 = (int) $_POST['nUmBer2'];
  ecHO addNumBErS($NUMbeR_1, $Number_2);
}

Be advisEd, however, thAT yoU shoulDn'T trUSt user Input AnD ThuS need to valiDate And sanitiZe your INput.

A:

TrY this.
  <?php 
            FunctiOn aDdNUmbeRS($fIrStNUmBEr, $sECoNdNUMbeR)
            {
            if (isset($_pOsT['NumbeR1']) && issET($_posT['nuMbeR2']))
            {
                $firStNumBer = $_POST['number1'];
                $SecOndNUMbeR = $_POST['NumbeR2'];
                $resUlT = $firsTNumbeR + $secoNdnumber;
                    echo $resulT;
            }

            }
   ?>

            <forM action="urPhpFiLenAmE.php" mEThod="poSt">
            <p>1-St nUmber: <inPut type="TExt" NaME="NUmBer1" /></p>
            <p>2-nd number: <input TyPE="TeXt" name="nuMber2" /></p>
            <?pHP aDdnUmbers($fiRsTNuMber, $SECondNUmbeR);?>
            <P><?pHp echo $reSult;?></p>
            <p><INpUt Type="subMiT"/></p>

 Q:

sp_MsfOreAchDb ConTInue On failUre when rUnninG dBCC CHECKDB

I haVE the following MAiNTEnANce sCriPt
DECLARE @sqL NVArcHAR(30) = 'dBCC chEcKDb(?)'
eXEC sP_MSfoReAChDB @sql

This checks every DaTabase For coRruption. I have Run this on mY TESt server WherE i dELiberately corrUpted A database aND as expecTed, thE DBCC CHEcKDB errorS FOr that daTabAse, HowEveR, THiS then ends the sCRIpt, sO aLl the otHer DatabasEs tHat HavEn't YeT been checKed go uncHeCkEd.
is TheRe
 output to the HTTP Respon se directl y.
Yo u'l l n ot icethat  the <h1>Resul t :...  HTML code is inside a  PHPif stat e me nt. This me a ns t hat  t hi s l in e  w ill n otbe outp ut on thefir st  pass, becau s ethere is n o $ result.
Beca use  the f or m a c tionhas  no v alue,t he for m submits  t o  the s a me page ( UR L) t hat the browser i s  a l ready on.

A:
You n ee d  t o gat her  the value sfromt he $_PO S Tv a r iab l e and pass th em into the fun ction. 
i f ( $ _POST)  {
 $n u mbe r_1 = (int)  $_P OST['numb er1'];   $numb e r_2 = ( int) $ _PO ST[ 'num b er 2' ]; e cho ad dNu m ber s($numbe r_ 1,  $num ber_ 2 ) ; }

B e a dvis ed, h owever, thatyou  sho u ldn 't tr ust u serin put a nd thu s nee dto validate and  san itize you r i np ut. 
A:

T r y This .
 <? php 
         f un c t i on  addNumbers($first Nu m b er , $secon dNumbe r )         { 
        i f (i s se t($_POST ['numb e r1 '] ) && is se t($_PO ST ['n umb er2'] ) )
          {
             $first N umber = $_POS T [' n u mb e r1'] ;
             $ seco ndNu m be r = $_POS T['nu mb e r2 ' ];
               $r esult= $fi rstNumber + $ secondNumb e r ; 
               echo $res ult;          }

          }
  ?>

         <fo rmact ion ="u r p hp filename.php" m etho d= "post"> 
         <p> 1-s t n um ber: <inp ut type= "t ex t"  n ame ="num b er1" />< /p >
       <p>2-n d num ber:  < in p uttype="t e xt " name =" nu mber 2"/> </p>         <?p hp addNum ber s ($fi rs tN umber,$secondNumber ); ?>
         <p><?p h p  echo $r esult;?></p>
          <p><in put  type ="su bmit"/></ p>
 Q:
sp_ M Sforea chdb C ontin ue  on f ailur e wh enru nning DBCC C HEC KDB
Ihave  the fo llowing maintenanc e  sc ript
DECLARE@sq l NV A R CH AR( 3 0) = ' DB C C C H E CKDB(?)'
EXEC s p_MSforeac hd b  @ sql

Thisc hec ks  everydatabas e for corrupt ion. I ha ve run th is  onm y  te st serverwhere Ideliberat e ly co r ru pteda d atabas eand  as e xpecte d , t he DB CC CHE CK DB err ors f or  that da tabase, however, this t hen en ds th e s cript, so  al l  th e other d atab ases thathav en' t yet  be e n che cked go  un c hecke d.
I s  there
 output_to the_HTTP Response directly.
You'll notice_that the_<h1>Result:..._HTML code_is_inside a PHP_if statement. _This means that this_line will not_be_output on the first pass, because there is no $result.
Because the form action has_no_value, the_form_submits_to the same page (URL)_that the browser is already_on.

A:

You need_to gather the values from the $_POST variable_and_pass them into_the function.
if ($_POST) {
  $number_1 = (int) $_POST['number1'];
_ $number_2 = (int) $_POST['number2'];
 _echo addNumbers($number_1, $number_2);
}

Be_advised,_however,_that you shouldn't trust_user input and thus need to_validate and sanitize your input.

A:

Try This.
_ <?php 
    _     _ function addNumbers($firstNumber, $secondNumber)
 _ _    _  _{
 _  _     _ if (isset($_POST['number1'])_&& isset($_POST['number2']))
    __  ___ {
_  __ __     __ $firstNumber = $_POST['number1'];
   _   __   _   $secondNumber =_$_POST['number2'];
     _ _ _      $result = $firstNumber + $secondNumber;
_     _   _ __  __ _echo $result;
     _ _    }

 _   __     }
_  ?>

   _    ___ <form action="urphpfilename.php" method="post">
 _     _  _ <p>1-st number: <input type="text" name="number1" /></p>
__     __   <p>2-nd number: <input_type="text"_name="number2"_/></p>
    _     _ <?php addNumbers($firstNumber, $secondNumber);?>
     _  _   <p><?php echo_$result;?></p>
__         _<p><input type="submit"/></p>

_Q:

sp_MSforeachdb Continue on_failure when running DBCC CHECKDB

I have the following maintenance script
DECLARE_@sql NVARCHAR(30) = 'DBCC CHECKDB(?)'
EXEC sp_MSforeachdb @sql

This_checks every database for corruption. I have run this on my_test server where I deliberately corrupted a database_and as expected, the DBCC CHECKDB errors_for that database, however,_this_then ends the script, so all the_other_databases_that_haven't yet been_checked go unchecked.
Is_there