Hello and thank you.
I have this code but, what where exactly should I compare the strings then if email was wrong just lift the info there but if right write it in the db.
Thank for any help.
[PHP]
<tr>
<td width="20%">Ema il:</td>
<td width="80%"><in put name="email" type="text" size=45 maxlength=80 value="<?if (isset($email))...
nwebhosting
Last Activity: Jan 21 '07, 08:23 AM
Joined: Oct 4 '06
Location:
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confirm previously entered email address
Hi.
I need help with this plzz:
Where I could find a php form script that confirms the email ( like enter email again just to make sure the user has enter the right email)
Thanks for any help. -
Thanks. You did help me.Leave a comment:
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image size
Hello.
Could you help me plzz?
I have an image table in a db and each image has different sizes.
When they get printted, some images look good some not.
Is there any way to adjust this code to make them print ok?.
Thanks for any help.
[HTML]
[PHP]
<tr>
<td width="60%"><im g border="1" height="124" width="164"... -
I do have the statement : $result = mysql_query ($sql, $conn);
Here is the code again
textimage20.php :
[HTML]
[PHP]
echo '<TD> <a href="http://localhost/text4.php?image _id=123"> <img border="1" height="90" width="100"
src="imagedolls .php?act=view&i id=' . $row['image_id'] . '"> </a> ';...Leave a comment:
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what is wrong with this code
I click on a image on textimage20.php
It takes me to text4.php but it all displays is:
SELECT gameandtoyname, pricevalue, bigdescription, dollsimage.imag e_id, dollsimage.imag e_date, doll.pricetype, prices.pricetyp e, prices.priceval ue FROM dollsimage, doll, prices WHERE doll.pricetype = prices.pricetyp e and dollsimage.imag e_id = '123' and doll.gameandtoy name = '123' and doll.bigdescrip tion = '123'
Then I see the... -
Sorry sir.
line 163 is exactly where I open <table>.
Thanks.Leave a comment:
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table
Dear sir.
I am trying to put my information this way:
gameandtoyname
pricevalue
image
but i am getting this error
Parse error: parse error in C:s\textimage10 .php on line 163
Could you help plzz.
[HTML]
[PHP]
<?php
<table>
<tr>';
<td>'$str .= "<a
href=\"textimag e10.php?iid=".$ row["image_id"]."\">".$row["image_name "]."</a>... -
Thanks alot. You did help me....Leave a comment:
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Sir here is the imagedolls.php:
Thanks
[PHP]
<?php
// database connection
$conn = mysql_connect(" localhost", "root") OR DIE (mysql_error()) ;
@mysql_select_d b ("products", $conn) OR DIE (mysql_error()) ;
$sql = "SELECT * FROM dollsimage WHERE image_id=".$_GE T["iid"];
$result = mysql_query ($sql, $conn);...Leave a comment:
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Thank you. It fixed the SELECT....Leave a comment:
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Printing image.
PHP Version:4.3.9 win32
Problem: I try to print the image from table dollsimage , gameandtoyname and the pricetype from dolls. (i have 7 images and 7 gameandtoyname) , but This code prints only 3 borders.
What i am doing wrong? i think the problem is with the $str.
Pleas help!
[PHP]
<?php
$sql = ("SELECT gameandtoyname, pricetype, dollsimage.imag e_id, dollsimage.imag e_date... -
[SOLVED] Mysql - Select
Hello and thank you for any help.
I have being trying to fix this problem reading diferent tutoriales but it has not being possible. I must be missing something in the code. Thankx
I get this error:
Invalid query: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE dollsimage.imag e_id = dolls.image_id' at line 4... -
Sir. now it brings this error:
thankx alot.
Invalid query: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM dollsimage, dolls ORDER BY image_date DESC...Leave a comment:
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Sir: I do have the connection. here is the rest of the code:
thankx alot.
[PHP]
<?php
// database connection
$conn = mysql_connect(" localhost", "root") OR DIE (mysql_error()) ;
@mysql_select_d b ("products", $conn) OR DIE (mysql_error()) ;
// browse the file and click the submit button
if ($_FILES) {
$image_types = Array...Leave a comment:
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Sir, still i get this error:
Warning: mysql_num_rows( ): supplied argument is not a valid MySQL result resource in C: on line 67
Could you please help me with this. Thanx
[PHP]
<?php
$sql = ("SELECT gameandtoyname, pricetype, dollsimage.imag e_id, dollsimage.imag e_date,
FROM dollsimage, dolls
ORDER BY image_date DESC
...Leave a comment:
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Sorry now is into tags
[PHP]
<?php
$sql = ("SELECT gameandtoyname, pricetype
* FROM dollsimage, dolls
ORDER BY image_date DESC
where dollsimage.imag e_id= dolls.image_id" );
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result)>0) {
while ($row = mysql_fetch_arr ay($result, MYSQL_ASSOC)) {
$i++;
...Leave a comment:
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Sorry Sir. I understand the php tags are: <?php ?>
sorry again.
Still can yuo help Me?....?
<?php
$sql = ("SELECT gameandtoyname, pricetype
* FROM dollsimage, dolls
ORDER BY image_date DESC
where dollsimage.imag e_id= dolls.image_id" );
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result)>0) {
...Leave a comment:
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<?php
$sql = ("SELECT gameandtoyname, pricetype
* FROM dollsimage, dolls
ORDER BY image_date DESC
where dollsimage.imag e_id= dolls.image_id" );
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result)>0) {
while ($row = mysql_fetch_arr ay($result, MYSQL_ASSOC)) {
$i++;
$str .= $i.". ";
...Leave a comment:
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Trying this way, it give me this error:
Warning: mysql_num_rows( ): supplied argument is not a valid MySQL result resource in on line 57
it is because i am missing something in the query?...Leave a comment:
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