php and random sql search

[smithandson]

Hello people,

 

I need some help with some PHP coding, and hope someone can help me.

 

I have a mysql table within a mysql database.

for example

 

id  word          sentence                                      associated word

1  happy      there is a lot of unhappiness                  unhappiness

2  consume  I am a consumer                                  consumer

3

 

 

I want to choose a random element from the associated word column

 

I want to make this random element a php variable  $word1 (for example)

 

elsewhere I want to have an text input box with a submit button

I want to make what is typed a php variable, for example $word 2

 

I then want to compare $word1 and $word2

 

If they are the same, I want to echo $word2 in green

If not, I want to echo $word2 in red with an explanation

 

I hope my description is not too confusing

 

many thanks

[Jessica]

What do you have so far? We can help debug the code but you should write it yourself. You already wrote the logic, now just convert it to code.

[smithandson]

thank you very much for such a fast response

 

What I have till now is as below

 

 

<?php

$con = mysql_connect ("localhost", "root", "root");

if (!$con)

  {

 

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

 

}

 

mysql_select_db("clients", $con);

 

$range_result = mysql_query( " SELECT MAX(`wd_id`) AS max_id , MIN(`wd_id`) AS min_id FROM `words` ");

 

$range_row = mysql_fetch_object( $range_result );

 

$random = mt_rand( $range_row->min_id , $range_row->max_id );

$result = mysql_query( " SELECT * FROM `words` WHERE `wd_id` >= $random LIMIT 0,1 ");

 

$answer = mysql_query("SELECT answer FROM words ORDER BY RAND() LIMIT 1");

 

while($row = mysql_fetch_array($answer))

  {

 

echo $row['answer'];

 

echo "<br />";

 

}

 

mysql_close($con);

?>

 

<?php

 

 

 

   

while($row = mysql_fetch_array($result))

  {

 

echo "<table border ='0' width='600px'>";

 

echo "<tr>";

 

echo "<td width='100px'> <font color='black'>

Clue word: </font></td>";

 

echo "<td width='100px'>" . $row['wd'] . "</td>";

 

echo "<td width='150px'> <font color='black'>Context sentence: </font></td>";

 

echo "<td width='200px'>" . $row['st'] . "</td>";

 

echo "<td>" . $row ['wd_id'] . "</td>";

 

echo "<td>" . $row ['answer'] . "</td>";

echo "</tr>";

}

 

echo"</table>";

 

  ?>

 

<?php

 

 

$input = $_POST['1'];

 

 

 

 

if (isset ($_POST['SUBMIT'])){

 

if ($input == $answer)

 

 

echo  "<font color='green'>" . $answer . "</font> ";

 

else echo "<font color='red'>" . $input . "</font>";

 

echo $explanation1;

 

echo $related = $row['related'];

 

echo $pron = $row['pron'];

 

 

}

 

?>

 

Please advise

 

kind regards

[Jessica]

1. Use code tags (The # button)

2. Tell us what doesn't work.

[smithandson]

Hi,

I can generate the random element with SQL, but only as echo

 

elsewhere I can do the php comparison and echo green or red depending on whether the answer is right or wrong.

 

I am having problems establishing the random output of the sql as a php variable, with which to do the comparison.

 

many thanks

[smithandson]

Hi

 

I think I need to be a bit clearer

 

I am starting with the following table

 

 

id word         sentance                                     answer               explanation

1 happy I am sorry for your unhappines     unhappiness

2 consume I am a consumer                     consumer

3

 

 

From this table I am  trying to  create the variable $answer which captures in the random row generated , the content in the column "answer" of that row. (for example 'unhappiness)

 

the code I have for this operation is:

 

 

<?php

 

$con = mysql_connect ("localhost", "root", "root");

if (!$con)

  {

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

  }

 

mysql_select_db("clients", $con);

$range_result = mysql_query( " SELECT MAX(`wd_id`) AS max_id , MIN(`wd_id`) AS min_id FROM `words` ");

$range_row = mysql_fetch_object( $range_result );

$random = mt_rand( $range_row->min_id , $range_row->max_id );

$result = mysql_query( " SELECT * FROM `words` WHERE `wd_id` >= $random LIMIT 0,1 ");

 

mysql_close($con);

?>

 

to echo this random answer, the code I have is

 

 

<?php

 

    while($row = mysql_fetch_array($result))

  {

  echo "<table border ='0' width='600px'>";

echo "<tr>";

echo "<td width='100px'> <font color='black'> Clue word: </font></td>";

echo "<td width='100px'>" . $row['wd'] . "</td>";

echo "<td width='150px'> <font color='black'>Context sentence: </font></td>";

echo "<td width='200px'>" . $row['st'] . "</td>";

echo "<td>" . $row ['wd_id'] . "</td>";

echo "<td>" . $row ['answer'] . "</td>";

echo "</tr>";

}

echo"</table>";

 

  ?>

 

I then have a text input box with a submit button

a word is typed in this input box which I wish to compare with the random answer previously generated and if they are the same, echo in green if not echo in red

 

the code for that is

 

 

<?php

 

$input = $_POST['1'];

 

if (isset ($_POST['SUBMIT'])){

 

if ($input == $answer)

 

echo  "<font color='green'>" . $answer . "</font> ";

else echo "<font color='red'>" . $input . "</font>";

echo $explanation;

echo $related;

echo $pron;

 

}

?>

 

the problem I have is that

 

when you try comparing the input from the form to the variable $answer, it doesn't compare.

 

Basically because I am not able to define to variable $anwer with the information I want it to have.

 

I need the code that defines the variable $answer (which would be content in  the columna "answer" of the random row generated by the variable $result).

 

any help would be very appreciated

 

many thanks

 

 

 

 

[Jessica]

$answer = $result['answer'];

 

I don't get why if you can echo it, you can't assign it....

[smithandson]

Hi,

 

Do you think my code is correct?

 

thanks

[smithandson]

Thank you very much for your help JESIROSE

 

you have been very kind.

 

I will be away for a bit and as such will not be able to try your solution.

 

As soon as I get it working I will give you some feedback.

 

thank you once againg, and take care

 

regards