[Values in array]

Thank you I think its the array_unique function.

 

I'm sorry for asking such a obvious question , I'm dyslexic so sometimes I miss things that are right in front of me.

 

Thanks

 

James

[Values in array]

array (james, john, pingo, adam, john, james, pingo, barney, pingo, );

 

I was wondering if there is an in-built PHP function that would list all the names once.

 

IE

 

james

john

pingo

adam

barney

 

not :

 

james

john

pingo

adam

john

james

pingo

barney

pingo

 

I have been googling this and cant seem to fnd what I want, having a hard time explain it in a google search term.

 

Thanks

 

James

[How Do I Show Numbering To Table?]

In your opinion of course.

 

yes in my opinion of course.

[How Do I Show Numbering To Table?]

$rowNo = 0;
while($res = mysql_fetch_array($result))
{
$rowNo++; //Increment Row Number
$date = date("D, j M, Y", $res['time']);
$center = strtoupper($res['center']);


if($res['approval'] == "Yes") {
$approvalText = "Approved":
$approvalColor = '#CDE861';
} else if($res['approval'] == "No") {
$approvalText = "Rejected":
$approvalColor = '#F28598';
} else {
$approvalText = "Pending":
$approvalColor = '#EDA553';
}?>


<tr align='center' bgcolor='<?php {$approvalColor} \n ?>'>
<td style='color:black;'><?php {$rowNo}; ?></td>
<td style='color:black;'><?php {$res['randid']}; ?></td>
<td style='color:black;'><?php {$center}; ?></td>
<td style='color:black;'><?php {$res['customer_name']}; ?></td>
<td style='color:black;'><?php {$res['home_phone']}; ?></td>
<td style='color:black;'><?php {$date}; ?></td>
<td><?php {$approvalText}; ?></td>
<td><a href ="view.php?id=<?php {$res['id']}; ?>"><center>Click to View </center></a></td><br />


//And continue on for other rows


</tr><br /><?php
}

 

sorry but thats the better way to right it instead of using all the echo codes

[How Do I Show Numbering To Table?]

<?php echo $i++ ;?>

[Php Updating Local Database]

can you give us more information on the project, like the code you already have and the fields in the database, also what ce do you current have. you need to tell us what you have what you intended it to do and what you want it to do.

[How To Shuffle An Array Then Implode It?]

it looks like its returning a boolean.

[Php Array Emailing Issues]

also instead of print_r try var_dump

 

Thats should give you more information and remember to use <pre> tags

[Building A Registration Form With Array And Foreach.]

the process you need to go through is input the the $_POST fields into an array.

 

$varible1 = $_POST['first_name'];

$varible2 = $_POST['last_name'];

$varible3 = $_POST['company'];

$varible4 = $_POST['email'];

 

$array = array($varible1, $varible2, $varible3, $varible4,);

 

foreach($array as $B){

 

$c[] = str_replace(' ', '_', $B);

 

 

}

 

 

this will gie you a formatted array.

[Can't Understand Why This Code is'nt Working]

Hi thanks for that

 

I replace that li with your code,

 

Now im getting an error on line 20

 

it makes no sence

 

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
if (!isset($_POST['id'])){
mysql_select_db("product", $con);

$result = mysql_query("SELECT * FROM  product2");

echo'<form method="post" action="pro.php">';
while($row = mysql_fetch_array($result))
  {
echo '<tr><td><input type="checkbox" name="uid[]" value="'.$row['ID'].'" /></td><td>"'.$row['filename'].'"</td><td>"'.$row['title'].'"</tr>';
}

echo '<input type="submit" name="formSubmit" value="Submit" /></form>';
?>

 

Thanks

 

James

[Can't Understand Why This Code is'nt Working]

ok here is the new "improved" code.

 

Also I'm getting a parse error on line 16

 

 

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
if (!isset($_POST['id'])){
mysql_select_db("product", $con);

$result = mysql_query("SELECT * FROM  product2");

echo'<form method="post" action="pro.php">';
while($row = mysql_fetch_array($result))
  {
  echo '<tr><td><input type="checkbox" name="uid[]" value=".$row['ID']." /></td><td>".$row['filename']."</td><td>".$row['title']."</tr>';
  }

echo '<input type="submit" name="formSubmit" value="Submit" /></form>';
?>

[Can't Understand Why This Code is'nt Working]

ok this is what I have but for some reason its stoped working geting a parse error on the last line 

 

 

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
if (!isset($_POST['id'])){
mysql_select_db("product", $con);

$result = mysql_query("SELECT * FROM  product2");

echo'<form method="post" action="pro.php">';
while($row = mysql_fetch_array($result))
  {
  echo "<tr><td><input type=\"checkbox\" name=\"uid[]\" value=\"".$row['ID']."\" /></td><td>".$row['filename'] . "</td><td> " . $row['title']. "</tr> ";
  }

echo '<input type=\"submit\" name=\"formSubmit\" value=\"Submit\" />';
echo '</form>';
mysql_close($con);
?>

[Can't Understand Why This Code is'nt Working]

hmmm now its only returning a single result

 

I have a group checklist HTML form that were the array is started. I wanted the script to pull details from the database when I select certain items. now if I sect multiple items it only returns one result. 

[Can't Understand Why This Code is'nt Working]

Hi

 

I cant understand why this code isnt echoing the results.

 

$ids = $_POST['uid'];

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

mysql_select_db("product", $con);

$sqlout = mysql_query("SELECT * FROM product2 WHERE 'ID' IN ('$ids')");

while ($sqlres = mysql_fetch_assoc($sqlout))
  {
  echo $sqlres['filename'] . " " . $sqlres['title'];
  }

mysql_close($con);

 

 

Anyone got any ideas?

 

Thanks

 

James

[slight sql problem]

Okay well did you try the code i posted?

 

Yes Chris, that solved tht problem but a new one has poped up. Thank you for your help its much appreciated

 

Thanks James 

[slight sql problem]

Hi

I made some adjustments to the code,

 

$ids = $_POST['uid'];

 

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

 

mysql_select_db("product", $con);

 

$sqlout = mysql_query("SELECT * FROM `product2` WHERE `ID` IN ('$ids')");

 

while ($sqlres = mysql_fetch_assoc($sqlout))

  {

  echo $sqlres['filename'] . " " . $sqlres['title'];

  }

 

mysql_close($con);

 

 

Im wondering why it isnt echoing the results?

 

Any thoughts

[slight sql problem]

here is were im setting the variable

 

$ids = $_POST['uid'];

[slight sql problem]

Hi

 

I have been getting this error.

 

mysql_fetch_array() expects parameter 1 to be resource, boolean given

 

Its driving me mad I know its something really simple. Can anyone see what it is?

 

mysql_select_db("product", $con);

$sqlout = mysql_query("SELECT * FROM 'tblt' WHERE 'UID' =('$ids')");

while($sqlres = mysql_fetch_array($sqlout))

 

 

Thanks

 

James

[Database ARRAY]

Why dont you just serialize the array It would be simplier and remember to deserial it when extracting it.

 

 

[strpos]

Hi Cunoodle

 

For some reason its starte working now I been playing wit it for a hour or so. I think the problem was i was getting the data from a file and looking for several tags all at once.

I just split them up a bit and thats seems to work

 

Thanks SO much for you help

 

James

 

[strpos]

snip echo out fine

 

in regards to finding just "<div id=" there several div tages in the code would it end up giving the position off all of them.

 

thanks

 

James 

[strpos]

Hi

Im having a problem finding the posistion of a div tag I tried this piece of code

 

$fpos= strpos($snip, "<div id=\"footer\">")

 

Then this piece

 

$fpos= strpos($snip, "<div id="footer">")

 

The $snip is the entire html document

 

I thought maybe the \ might be causeing problem but without it

 

Then i tried this

 

any suggestions guys

 

Thanks

 

James

[problem with Query]

im still no futher on this if anyone could help

[problem with Query]

ALso i ran a check and it not coming up as a object or an array

[problem with Query]

Hi

 

Im just working on a wordpress pugin at the moment and Iw as trying out this piece of code

 

$mylink = $wpdb->get_row("SELECT * FROM $wpdb->testtable WHERE id = 1");

 

echo $mylink->id ;

 

there is data in the table there is an id equal to one but it not coming up with anything

 

I tried it with anouther pable and it works fine I can for example

 

$mylink = $wpdb->get_row("SELECT * FROM $wpdb->posts WHERE id = 1");

 

echo $mylink->post_content ;

 

That comes up fine with the data

 

But the first one I dont know why its not working

 

 

can anyone help

 

 

Thanks

 

James