print $str

[nwebhosting]

Hi. I am very new with php. can you help me with this:
This code is printing part of the string: image and image_id, but gameandtoyname and pricetype NO. What i am doing wrong? any help?

<?php
$sql = "SELECT gameandtoyname,pricetype * FROM dollsimage, dolls where dollsimage.image_id=dolls.image_id";
$sql = "SELECT * FROM dollsimage ORDER BY image_date DESC";
$result = mysql_query ($sql, $conn);
if (mysql_num_rows($result)>0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$str .= $i.". ";

$str .="<img border=\"1\" height=\"90\" width=\"100\" src=\"imagedolls.php?act=view&iid=".$row["image_id"]." ".$row["gameandtoyname"]." ".$row["pricetype"]." \"></a> ";

}
print $str;
}
?>

[ronverdonk]

Only the last query is executed, because you assigned the last SELECT string to the same variable ($sql) as the first SELECT string (also in variable $sql).

Ronald :cool:

[nwebhosting]

Only the last query is executed, because you assigned the last SELECT string to the same variable ($sql) as the first SELECT string (also in variable $sql).

Ronald :cool:

Hi and thanks.

If i use 2 differentes variables for SELECT, i will have to use 2 differents variables for the query. How could i use only one for the same condition and print only one string?

Please help me! it is tricky and i have being using different thing but it give me errors.

[ronverdonk]

Combine the 2 queries into one, selecting columns from both tables in 1 SELECT statement. Such as in this sample:

SELECT fielda, fieldb, fieldc, fieldd 
FROM table_1, table_2 
WHERE ......

Ronald :cool:

[nwebhosting]

Combine the 2 queries into one, selecting columns from both tables in 1 SELECT statement. Such as in this sample:

SELECT fielda, fieldb, fieldc, fieldd 
FROM table_1, table_2 
WHERE ......

Ronald :cool:

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?

[ronverdonk]

Show the code regarding the query and the processing of the results WITHIN [PHP] TAGS!

Ronald :cool:

[nwebhosting]

Show the code regarding the query and the processing of the results WITHIN [PHP] TAGS!

Ronald :cool:

<?php
$sql = ("SELECT gameandtoyname,pricetype
* FROM dollsimage, dolls
ORDER BY image_date DESC
where dollsimage.image_id= dolls.image_id");

$result = mysql_query ($sql, $conn);
if (mysql_num_rows($result)>0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$str .= $i.". ";

$str .="<img border=\"1\" height=\"90\" width=\"100\" src=\"imagedolls.php?act=view&iid=".$row["image_id"]."".$row["gameandtoyname"]."".$row["pricetype"]." \"></a> ";

}
print $str;
}
?>

[ronverdonk]

Now I explicitely requested you to put your code between tags!!
Obviously you refuse to fulfill such a simple request, even though it is in the POSTING GUIDELINES!!

Therefore I feel no great urge to continue this discussion any further.

Ronald :cool:

[nwebhosting]

Now I explicitely requested you to put your code between tags!!
Obviously you refuse to fulfill such a simple request, even though it is in the POSTING GUIDELINES!!

Therefore I feel no great urge to continue this discussion any further.

Ronald :cool:

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.image_id= dolls.image_id");

$result = mysql_query ($sql, $conn);
if (mysql_num_rows($result)>0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$str .= $i.". ";

$str .="<img border=\"1\" height=\"90\" width=\"100\" src=\"imagedolls.php?act=view&iid=".$row["image_id"]."".$row["gameandtoyname"]."".$row["pricetype"]." \"></a> ";

}
print $str;
}
?>

[nwebhosting]

Sorry now is into tags

[PHP]
<?php
$sql = ("SELECT gameandtoyname,pricetype
* FROM dollsimage, dolls
ORDER BY image_date DESC
where dollsimage.image_id= dolls.image_id");

$result = mysql_query ($sql, $conn);
if (mysql_num_rows($result)>0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$str .= $i.". ";

$str .="<img border=\"1\" height=\"90\" width=\"100\" src=\"imagedolls.php?act=view&iid=".$row["image_id"]."".$row["gameandtoyname"]."".$row["pricetype"]." \"></a> ";

}
print $str;
}
?>

[/PHP]

[ronverdonk]

Ok. Just to make it easier to read, I have assigned the $row values to separate php variables, so it is easier to include in the output string. For the same reason I have also replaced the \"by a simple single slash. That is no criticism, just makes it somewhat easier to read.

Now what was wrong was, that you had in your select statement an * without separating comma, so there was an sql error.
Also I saw that you forgot to include the field image_date in your select, so I added that. I do not know from which table you want to select image_date, but I assumed from table dollsimage, so you'll have yo change that if not correct.

That brings me to the following code snippet for you.[php]<?php
$sql = ("SELECT gameandtoyname, pricetype, dollsimage.image_id, dollsimage.image_date,
FROM dollsimage, dolls
ORDER BY image_date DESC
WHERE dollsimage.image_id = dolls.image_id");

$result = mysql_query ($sql, $conn);
$str = "";
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$image_id = $row["image_id"];
$gameandtoyname = $row["gameandtoyname"];
$pricetype = $row["pricetype"];
$str .="$i. <img border='1' height='90' width='100' src='imagedolls.php?act=view&iid=$image_id$gameand toyname$pricetype'></a> ";
print $str;
}
print $str;
}
?>[/php]
Let me know if this works (after you checked that the image_date is selected from the correct table).

Ronald :cool:

[nwebhosting]

Ok. Just to make it easier to read, I have assigned the $row values to separate php variables, so it is easier to include in the output string. For the same reason I have also replaced the \"by a simple single slash. That is no criticism, just makes it somewhat easier to read.

Now what was wrong was, that you had in your select statement an * without separating comma, so there was an sql error.
Also I saw that you forgot to include the field image_date in your select, so I added that. I do not know from which table you want to select image_date, but I assumed from table dollsimage, so you'll have yo change that if not correct.

That brings me to the following code snippet for you.[php]<?php
$sql = ("SELECT gameandtoyname, pricetype, dollsimage.image_id, dollsimage.image_date,
FROM dollsimage, dolls
ORDER BY image_date DESC
WHERE dollsimage.image_id = dolls.image_id");

$result = mysql_query ($sql, $conn);
$str = "";
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$image_id = $row["image_id"];
$gameandtoyname = $row["gameandtoyname"];
$pricetype = $row["pricetype"];
$str .="$i. <img border='1' height='90' width='100' src='imagedolls.php?act=view&iid=$image_id$gameand toyname$pricetype'></a> ";
print $str;
}
print $str;
}
?>[/php]
Let me know if this works (after you checked that the image_date is selected from the correct table).

Ronald :cool:

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.image_id, dollsimage.image_date,
FROM dollsimage, dolls
ORDER BY image_date DESC
WHERE dollsimage.image_id = dolls.image_id");

$result = mysql_query ($sql, $conn);
$str = "";
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$image_id = $row["image_id"];
$gameandtoyname = $row["gameandtoyname"];
$pricetype = $row["pricetype"];
$str .="$i. <img border='1' height='90' width='100' src='imagedolls.php?act=view&iid=$image_id$gameand toyname$pricetype'></a> ";
print $str;
}
print $str;
}



?>

[/PHP]

[ronverdonk]

Of course that code does not work, Before it you'll have to connect to your server and your database. Such as:

[php]
$con = mysql_connect("xx", "yy", "zz")
or die("Connect error: " . mysql_error());
mysql_select_db("dd")
or die("Db select error: " . mysql_error());
[/php]
Don't forget to change the xx, yy, zz and dd to the actual values!

After that you put the select code.

Ronald :cool:

[nwebhosting]

Of course that code does not work, Before it you'll have to connect to your server and your database. Such as:

[php]
$con = mysql_connect("xx", "yy", "zz")
or die("Connect error: " . mysql_error());
mysql_select_db("dd")
or die("Db select error: " . mysql_error());
[/php]
Don't forget to change the xx, yy, zz and dd to the actual values!

After that you put the select code.

Ronald :cool:

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_db ("products", $conn) OR DIE (mysql_error());

// browse the file and click the submit button
if ($_FILES) {
$image_types = Array ("image/bmp",
"image/jpeg",
"image/pjpeg",
"image/gif",
"image/x-png");

$userfile = addslashes (fread (fopen ($_FILES["userfile"]["tmp_name"], "r"), filesize ($_FILES["userfile"]["tmp_name"])));
$file_name = $_FILES["userfile"]["name"];
$file_size = $_FILES["userfile"]["size"];
$file_type = $_FILES["userfile"]["type"];

if (in_array (strtolower ($file_type), $image_types)) {
$sql = "INSERT INTO dollsimage (image_type, image, image_size, image_name, image_date) ";
$sql.= "VALUES (";
$sql.= "'{$file_type}', '{$userfile}', '{$file_size}', '{$file_name}', NOW())";
@mysql_query ($sql, $conn);
Header("Location:".$_SERVER["PHP_SELF"]);
exit();
}
}

// view or remove
if ($_GET) {
$iid = $_GET["iid"];
$act = $_GET["act"];
switch ($act) {
case rem:
$sql = "DELETE FROM dollsimage WHERE image_id=$iid";
@mysql_query ($sql, $conn);
Header("Location:./textimage4.php");
exit();
break;
default:
print "<img src=\"imagedolls.php?iid=$iid\">";
break;
}
}

?>

<form method="post" enctype="multipart/form-data">
Select Image File: <input type="file" name="userfile" size="40"><input type="submit" value="submit">
</form>
<?php
$sql = ("SELECT gameandtoyname, pricetype, dollsimage.image_id, dollsimage.image_date,
FROM dollsimage, dolls
ORDER BY image_date DESC
WHERE dollsimage.image_id = dolls.image_id");

$result = mysql_query ($sql, $conn);
$str = "";
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$i++;
$image_id = $row["image_id"];
$gameandtoyname = $row["gameandtoyname"];
$pricetype = $row["pricetype"];
$str .="$i. <img border='1' height='90' width='100' src='imagedolls.php?act=view&iid=$image_id$gameand toyname$pricetype'></a> ";
print $str;
}
print $str;
}



?>
[/PHP]



Imagedolls.php :

[PHP]
<?php


// database connection
$conn = mysql_connect("localhost", "root") OR DIE (mysql_error());
@mysql_select_db ("products", $conn) OR DIE (mysql_error());
$sql = "SELECT * FROM dollsimage WHERE image_id=".$_GET["iid"];
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result)>0) {
$row = @mysql_fetch_array ($result);
$image_type = $row["image_type"];
$image = $row["image"];
Header ("Content-type: $image_type");
print $image;
}
?>
[/PHP]

[ronverdonk]

I cannot see the error, but it is late, so I could overlook it. You could add the following line after the query to see if it returns an error message:
[php]if (!$result) {
die('Invalid query: ' . mysql_error());
}[/php]
Ronald :cool:

[nwebhosting]

I cannot see the error, but it is late, so I could overlook it. You could add the following line after the query to see if it returns an error message:
[php]if (!$result) {
die('Invalid query: ' . mysql_error());
}[/php]
Ronald :cool:

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