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topshelfbleu

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Posts posted by topshelfbleu

    copying a table with a new column

    in MySQL Help

    Posted

    Hi

    I have an answers  table of 15 rows each with 100  fields (100) that I need to multiply by a factor of either +1 or -1 and then save the new version to a new table

     

    What is the best way to go about this?

     

     

    Any help really gratefully accepted.

     

    The concatenate command -

    in MySQL Help

    Posted

    Hi

    I have a table that looks like

     

    userid  qnum1  qnum2 qnum 3

    1              a          b        c

    2              d            a        f

     

    I want to create a new table using this data but organised in a different format  i.e.

    Answers

    "userid","columnamefrom table1", "answer"

     

    So this would look like

     

    Answers

    1;1;a

    1;2;b

    1;3;c

    2;1;d

    2;2;a

    2;3;f

     

    Is concatenate the way to go here?

     

     

     

     

     

     

     

    the IF command. Running a qry if a condition is met.

    in PHP Coding Help

    Posted

    Hi

    I've got a qry that selects 100 questions one at a time from tblquestions.

    When the users have answered q100 I want them to be able to come out of the post-process-post loop that has taken them through the survey.

     

    My idea is that I could put an IF requirement in before the qry runs each time it selects the next question

     

    i.e.

     

    if $nextq <101
  $result = mysql_query("SELECT oid,psc9 FROM tblquestions WHERE oid = $nextq");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
};
else echo "thanks you've finished now";

     

    Two Questions:

    • Is this the way I should be going about this problem?
    • Have I got the syntax right?

     

    Many thanks!

    Nick

     

     

    Problem with echoing some vars

    in PHP Coding Help

    Posted

    Hi

    I am attempting to request a single row from a table that has 20 fields.

    I want to turn each field into it's own variable (you'll see oc1, oc2 etc in the code) and then echo the variable just to check it has worked.

     

    When I run the qry below - it picks up the first one correctly but then places the same value in all the other fields. (The $question var should be a long text string but is echoing the first variable instead.

    There's clearly something wrong with the way I have constructed this - but I don't know at which point it has gone wrong.

     

    Can anyone help?

     

    $result = mysql_query("SELECT oid,qid,question,posneg,oc1,oc2,oc3,oc4,oc5,oc6,oc7,psc1,psc2,psc3,psc4,psc5,psc6,psc7,psc8,psc9 FROM tblquestions WHERE oid = '1'");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}

$oid=mysql_result($result,"oid");
$qid=mysql_result($result,"qid");
$question=mysql_result($result,"question");
$posneg=mysql_result($result,"posneg");
$oc1=mysql_result($result,"oc1");
$oc2=mysql_result($result,"oc2");
$oc3=mysql_result($result,"oc3");
$oc4=mysql_result($result,"oc4");
$oc5=mysql_result($result,"oc5");
$oc6=mysql_result($result,"oc6");
$oc7=mysql_result($result,"oc7");
$psc1=mysql_result($result,"psc1");
$psc2=mysql_result($result,"psc2");
$psc3=mysql_result($result,"psc3");
$psc4=mysql_result($result,"psc4");
$psc5=mysql_result($result,"psc5");
$psc6=mysql_result($result,"psc6");
$psc7=mysql_result($result,"psc7");
$psc8=mysql_result($result,"psc8");
$psc9=mysql_result($result,"psc9");
?>
  
<?php echo $oid; // Question Number?>
<?php echo $qid; // Question Number?>
<?php echo $question; // Question Number?>
<?php echo $posneg; // Question Number?>
<?php echo $oc1; // Question Number?>
<?php echo $oc2; // Question Number?>
<?php echo $oc3; // Question Number?>
<?php echo $oc4; // Question Number?>
<?php echo $oc5; // Question Number?>
<?php echo $oc6; // Question Number?>
<?php echo $oc7; // Question Number?>
<?php echo $psc1; // Question Number?>
<?php echo $psc2; // Question Number?>
<?php echo $psc3; // Question Number?>
<?php echo $psc4; // Question Number?>
<?php echo $psc5; // Question Number?>
<?php echo $psc6; // Question Number?>
<?php echo $psc7; // Question Number?>
<?php echo $psc8; // Question Number?>
<?php echo $psc9; // Question Number?>

    variable creation from table - Beginners question

    in PHP Coding Help

    Posted

    Hi -

    I know this is a #101 question but here goes!

     

    I simply want to select  a single  row from a table and use the columns fields as variables that can then be posted in a form.

     

    $result = mysql_query("SELECT * from tblquestions WHERE qid='1'");


if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
$row = mysql_fetch_row($result);

$oid=$result['oid'];
$qid=$result['qid'];
$question=$result['question'];

     

     

    The code above isn't working - as when I try to echo each variable 

    <?php echo $question;?>

    nothing is being shown on screen.

     

    I'm obviously barking up the wrong tree in the way I'm trying to create the variable - but don't really have any idea where to look next.

    Posting Variable in a hidden field.

    in PHP Coding Help

    Posted

    I am retrieving a single row from a table and I want to use one of the fields in a post form which will be sent to process.php

    I have successfully use echo to place some results where I want them – but I want to use some of the row results as hidden fields.

     

    The code I am using to retrieve from the table is 

    $result = mysql_query("SELECT qid,question,oc1 FROM tblquestions WHERE qid = '1'");

     

    The oc1 field  is the one I want to post by way of a hidden field so that it can be passed to the process.php.

     

    I have tried

    [/echo $row[2]; // oc1
$oc1 = $row[2];php]

And then in the form:

[code=php:0]<input type="hidden" name="oc1" id="oc1" value= <?php echo $oc1?> />

     

    I think I just read that this can't actually be done- but I'm new to all this, so am not sure whether I've understood how I should be doing it instead.

     

    Any help REALLY gratefully received! ;)

     

     

     

     

    Form inserting zeros into table

    in PHP Coding Help

    Posted

    I have a form that is posting a value from a radio button as well as 16 Hidden values.

    The form action goes to file2- which is successfully conecting and processing but instead of inserting the correct values I'm just seeing a row of zeros.

     

    I don't know where to start looking!

     

    This is the code from the posting form :

    ---

     

    <?php // query.php
require_once 'login.php';
$db_server = mysql_connect($db_hostname, $db_username, $db_password);

if (!$db_server) die("Unable to connect to MySQL: " . mysql_error());

mysql_select_db($db_database)
    or die("Unable to select database: " . mysql_error());


$result = mysql_query("SELECT qid,question,oc1,oc2,oc3,oc4,oc5,oc6,oc7,psc1,psc2,psc3,psc4,psc5,psc6,psc7,psc8,psc9 FROM tblquestions WHERE qid = '1'");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
$row = mysql_fetch_row($result);

echo $row[0]; // Question Number
echo $row[1]; // Question
echo $row[2]; // oc1
echo $row[3]; // oc2
echo $row[4]; // oc3
echo $row[5]; // oc4
echo $row[6]; // oc5
echo $row[7]; // oc6
echo $row[8]; // oc7
echo $row[9]; // pcs1
echo $row[10]; // pcs2
echo $row[11]; // pcs3
echo $row[12]; // pcs4
echo $row[13]; // pcs5
echo $row[14]; // pcs6
echo $row[15]; // pcs7
echo $row[16]; // pcs8
echo $row[17]; // pcs9


?>
  <p>Hello </p>
<?php 

echo $row[0]; // Question Number

?>
  <form id="form1" name="form1" method="post" action="./uinputprocess.php">
    <table width="800" border="0">
      <tr>
        <td colspan="6"><p>Question
          <?php 

echo $row[0]; // Question Number

?>
        </p>
        <?php 

echo $row[1]; // Question

?>
        .</td>
      </tr>
      <tr>
        <td width="126"><p align="center"> </p>
        <p align="center">Strongly Disagree        </p></td>
        <td width="65"><p align="center"> </p>
        <p align="center">Disagree        </p></td>
        <td width="145"><p align="center"> </p>
        <p align="center">Unsure but Doubtful        </p></td>
        <td width="152"><p align="center"> </p>
        <p align="center">Unsure but Probable        </p></td>
        <td width="67"><p align="center"> </p>
        <p align="center">Agree        </p></td>
        <td width="219"><p align="center"> </p>
        <p align="center">Strongly Agree        </p></td>
      </tr>
      <tr>
        <td><div align="center">
          <input type= "radio" name= "ans" value = "1"  />
        </div></td>
        <td><div align="center">
          <input type= "radio" name= "ans" value = "2"  />
        </div></td>
        <td><div align="center">
          <input type= "radio" name= "ans" value = "3"  />
        </div></td>
        <td><div align="center">
          <input type= "radio" name= "ans" value = "4"  />
        </div></td>
        <td><div align="center">
          <input type= "radio" name= "ans" value = "5"  />
        </div></td>
        <td><div align="center">
          <input type= "radio" name= "ans" value = "6"  />
        </div></td>
      </tr>
      <tr>
        <td colspan="6"> </td>
      </tr>
      <tr>
        <td colspan="6"><p>Hidden Fields Here > 
          <input type="hidden" name="oc1" id= <?php 

"echo $row[2];"

?> />   <input type="hidden" name="oc2" id= <?php 

"echo $row[3];"

?> />   <input type="hidden" name="oc3" id= <?php 

"echo $row[4];"

?> />   <input type="hidden" name="oc4" id= <?php 

"echo $row[5];"

?> />   <input type="hidden" name="oc5" id= <?php 

"echo $row[6];"

?> />   <input type="hidden" name="oc6" id= <?php 

"echo $row[7];"

?> />   <input type="hidden" name="oc7" id= <?php 

"echo $row[8];"

?> />   <input type="hidden" name="psc1" id= <?php 

"echo $row[9];"

?> />   <input type="hidden" name="psc2" id= <?php 

"echo $row[10];"

?> />   <input type="hidden" name="psc3" id= <?php 

"echo $row[11];"

?> />   <input type="hidden" name="psc4" id= <?php 

"echo $row[12];"

?> />   <input type="hidden" name="psc5" id= <?php 

"echo $row[13];"

?> />   <input type="hidden" name="psc6" id= <?php 

"echo $row[14];"

?> />   <input type="hidden" name="psc7" id= <?php 

"echo $row[15];"

?> />   <input type="hidden" name="psc8" id= <?php 

"echo $row[16];"

?> />   <input type="hidden" name="psc9" id= <?php 

"echo $row[17];"

?> />
        </p>
          <p>Pin 
            <label for="textfield"></label>
            <input name="textfield" type="password" id="textfield" size="20" maxlength="5" />
          </p></td>
      </tr>
      <tr>
        <td colspan="6"><p>
          <input type="submit" name="button" id="button" value="Submit" />
        </p>
        <p> </p></td>
      </tr>
    </table>
  </form>

     

    And this is the code from the processing file

     

     <?php // query.php
require_once 'login.php';
$db_server = mysql_connect($db_hostname, $db_username, $db_password);

if (!$db_server) die("Unable to connect to MySQL: " . mysql_error());

mysql_select_db($db_database)
    or die("Unable to select database: " . mysql_error());

$pin =$_POST['pin'];
$qid =$_POST['qid'];
$oc1 =$_POST['oc1'];
$oc2 =$_POST['oc2'];
$oc3 =$_POST['oc3'];
$oc4 =$_POST['oc4'];
$oc5 =$_POST['oc5'];
$oc6 =$_POST['oc6'];
$oc7 =$_POST['oc7'];
$psc1 =$_POST['psc1'];
$psc2 =$_POST['psc2'];
$psc3 =$_POST['psc3'];
$psc4 =$_POST['psc4'];
$psc4 =$_POST['psc5'];
$psc6 =$_POST['psc6'];
$psc7 =$_POST['psc7'];
$psc8 =$_POST['psc8'];
$psc9 =$_POST['psc9'];
$ans =$_POST['ans'];

$enter_sql= "INSERT INTO tblresults (pin,qid,oc1,oc2,oc3,oc4,oc5,oc6,oc7,psc1,psc2,psc3,psc4,psc5,psc6,psc7,psc8,psc9) VALUES ('$pin','$qid','$oc1','$oc2','$oc3','$oc4','$oc5','$oc6','$oc7','$psc1','$psc2','$psc3','$psc4','$psc5','$psc6','$psc7','$psc8','$psc9')";
$enter_query =mysql_query($enter_sql) or die (mysql_error());
?>
<body>
<p>Thank you - You have successfully entered your answer</p>

    import csv into via phpmyadmin

    in MySQL Help

    Posted

    Having trouble with this-

    Basically have created an excel csv and converted into notepad.

     

    Clicking Import in phpmyadmin - choosing CSV with Load Data and changed the ; to a comma.

     

    Clicking go I'm getting the following error :

     

    SQL query:

     

    LOAD DATA INFILE '/tmp/phpXqCUTK' INTO TABLE `tblquestions` FIELDS TERMINATED BY ';' ENCLOSED BY '"' ESCAPED BY '\\' LINES TERMINATED BY '\r\n'

     

    MySQL said: Documentation

    #1045 - Access denied for user 'myusername'@'%' (using password: YES)

     

    -

    This doesn't make any sense to me as phpmyadmin is doing everything else okay (with user password settings).

     

    Am I being totally stupid here?

     

    Carrying hidden fields in a form.

    in PHP Coding Help

    Posted

    Hi

    I've got a form that users enter an identification pin in for their first question  as well as clicking a radio button(Q1.php) . They click submit and the answers are put through a couple of multiplication function and then placed in a table (alongside the pin)  via Q2.php.

     

    On the Q2 page I want to place the 2nd question that the user needs to answer. 

     

    Rather than making them retype the pin I was hoping to process it as a hidden field - being picked up from their Q1.php entry.

    This 'hidden pin echo ' process will continue throughout Q2- Q10 pages.

     

    However - I can't seem to include any php in the page after the Q1 form has been processed.  Have I got my php tags messed up?

    <?php
$con = mysql_connect("localhost","ca","d");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("candango", $con);

$q1 =$_POST['q1'];
$pin =$_POST['pin'];


$ans1 = $q1 * 3;
$ans2 = $q1 * 5;
$ans3 = $q1 * 2;


$enter_sql= "INSERT INTO aapcm2 (pin,ans1, ans2, ans3)
VALUES ('$pin','$ans1','$ans2','$ans3')";
$enter_query =mysql_query($enter_sql) or die (mysql_error());
?>

<body>

<p>Thank you - You have successfully entered your answers.</p><form action="q2.php" method="post">
  <p>What do you think the answer to this one is? 
    <input type="radio" name="q2" id="q2" value="-3">
    <label for="q2"></label>
  <input name="Hidden" type="hidden" id="Hidden" value="$pin" /></p>
  <p>
    <input type="submit" name="button" id="button" value="Submit">
  </p>
</form>

<p></p>
</body>

    Multiplication of a variable - then entry into a table.

    in PHP Coding Help

    Posted

    Hi

    The page I'm working on will receive a value from a radio button ( '$q1' and then I need to apply 3 seperate multiplications to it. The 3 values will then be placed into a table.

     

    How do I turn the values into named variables which then can get placed into the dbase using the insert into dbasename  and values

     

    So far I've done this

     

    <?php

    $con = mysql_connect("blah","blah","blah");

    if (!$con)

      {

      die('Could not connect: ' . mysql_error());

      }

    mysql_select_db("candango", $con);

     

    $q1 =$_POST['q1'];

    $multiplication = '$q1'*3;

    $multiplication = '$q1'*5;

    $multiplication = '$q1'*10;

     

     

    $enter_sql= "INSERT INTO answers (answer from 1, answer from 2, answer 3)

     

    $enter_query =mysql_query($enter_sql) or die (mysql_error());

    ?>

    [/code]

     

    Any guidance - greatly accepted!  8)

     

    Thanks

     

    Nick

    Inserting specific mysql records into a php post form

    in PHP Coding Help

    Posted

    I want to pull some records  plus unique key from one mysql table and then use them as questions in a form/questionnaire.

     

    The post form will have six radio buttons  to the right of the fields pulled with the key posted but hidden and will wirte to another table.

     

    I guess this is a templating question- as I've successfully retrieved the table 1 records via a query - I just can't seem to use them in the form. Below is the code I'm using to view table "PCM1"

     

    <?php // query.php
require_once 'login.php';
$db_server = mysql_connect($db_hostname, $db_username, $db_password);

if (!$db_server) die("Unable to connect to MySQL: " . mysql_error());

mysql_select_db($db_database)
or die("Unable to select database: " . mysql_error());

$query = "SELECT * FROM pcm1";
$result = mysql_query($query);

if (!$result) die ("Database access failed: " . mysql_error());

$rows = mysql_num_rows($result);

for ($j = 0 ; $j < $rows ; ++$j)
{
echo 'RID: '   . mysql_result($result,$j,'mcde')  ;
echo 'Statement: '    . mysql_result($result,$j,'oppclub')    . '<br />';

;
}

    Posting into Mysql--am looking at a blank page.

    in PHP Coding Help

    Posted

    Yep - I do see the difference.... and thanks for pointing out that particular issue.

     

    However...

     

    I'm embarrassed to say I don't know what  PFMaBiSmAd's s quote

     

    "debugging php code on a system with error_reporting set to E_ALL and display_errors set to ON in your master php.ini so that php will help you" actually means. 

     

    :-[

     

     

    Posting into Mysql--am looking at a blank page.

    in PHP Coding Help

    Posted

    thanks all- but still nothing appearing.

     

    (I have corrected the straight bracket). I've been typing into dw rather than use a debugging tool. Am just having a look now at netbeans

     

    the post form is http://www.candango.co.uk/php/insert_records.php

     

    code for entry_page.php is

     

    <?php

$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("aaworldcup", $con);

$Name =$_POST['Name'];
$email =$_POST['email'];
$m1hscore =$_POST['m1hscore'];
$m1ascore =$_POST['m1ascore'];

$enter_sql= "INSERT INTO aaworldcup (Name,email,m1hscore,m1ascore) VALUES ('$Name','$email','$m1hscore','$m1ascore')";
$enter_query =mysql_query($enter)sql) or die (mysql_error());
?>
<body> hello</body>

    Posting into Mysql--am looking at a blank page.

    in PHP Coding Help

    Posted

    I've been looking at this for over an hour- I've even opened a book.

    I'm totally new to it and trying to write a form which allow friends to post predictions for each world cup match.

     

    Is there anything that stands out to you here. Once clicking submit - the retrieving page isn't showing anything and  obv not submitting to the dbase.

     

    <?php

$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("aaworldcup", $con);

$Name =$_POST['Name'];
$email =$_POST['email'};
$m1hscore =$_POST['m1hscore'];
$m1ascore =$_POST['m1ascore'];

$enter_sql= "INSERT INTO aaworldcup (Name,email,m1hscore,m1ascore) VALUES ('$Name','$email','$m1hscore','$m1ascore')";
$enter_query =mysql_query($enter)sql) or die (mysql_error());
?>

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