Php syntax error?

Question

eduardc 0 Light Poster

12 Years Ago

I don´t understand the error:

Parse error: syntax error, unexpected T_VARIABLE in /home/eduardli/public_html/web_designer/insert.php on line 21

php file:

<html>
<body>

<?php
$con = mysql_connect("localhost","eduardli_user","-z.x,c");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("eduardli_company", $con);


mysql_query ("INSERT INTO Products (Description, Price, Quantity)  
VALUES ('$description', '$price', '$quantity')");

mysql_query("SELECT*FROM Products='"$result"'");                  

while($row = mysql_fetch_array($result))
{
echo $row['Description'] . " " . $row['Price'] . " " . $row['Quantity'];
echo "<br />";
}

mysql_close($con);
?>

</body>
</html

form.html:

<html>
<head>
<style type="text/css">

#desc
{
width:10%;
height:5%;
}
</style>
<title>POST</title>
</head>

<body>


<form action='insert.php' method="post">

Description: <input type="text" name="description" id="description"><br />
Price: <input type="text" name="price" id="price" /><br />
Quantity: <input type="text" name="quantity" id="quantity" />
<input type='submit' value='submit' />


</form>



</body>
</html>

php

Edited 12 Years Ago by Ezzaral because: Added code tags. Please use them to format any code that you post.

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Latest Post 12 Years Ago Latest Post by karthik_ppts

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kekkaishi 18 Junior Poster

12 Years Ago

eduardc,
Your code looks to be OK. As Karthik suggested, try executing the line (select * from Products) in phpmyadmin (or your sql dbms).

Also, I suggest using mysql_error() in your codes to identify the exact error the database is returning. That is,

mysql_select_db("eduardli_company", $con) or die(mysql_error());


mysql_query ("INSERT INTO Products (Description, Price, Quantity) 
VALUES ('$description', '$price', '$quantity')") or die(mysql_error());

$q="select * from Products";
$result = mysql_query($q,$con) or die(mysql_error());

Regards.

karthik_ppts commented: useful post +3

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rpv_sen 59 Junior Poster · Answer 1 · 2011-05-02T09:52:22+00:00

Hi

In line number 21 the statement is

mysql_query("SELECT*FROM Products='"$result"'");

Please change as mysql_query("SELECT*FROM Products='$result'");

vibhaJ 126 Master Poster · Answer 2 · 2011-05-02T11:56:09+00:00

you need to replace this line.

mysql_query("SELECT*FROM Products='"$result"'");

to

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

karthik_ppts 81 Posting Pro · Answer 3 · 2011-05-02T12:46:39+00:00

karthik_ppts 81 Posting Pro

12 Years Ago

Use code tags to place your codes

eduardc 0 Light Poster · Answer 4 · 2011-05-02T19:07:20+00:00

Thanks! But unfornately this is´n the solution!

Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in /home/eduardli/public_html/web_designer/insert.php on line 23

P. s. I want to show the output in a diagram on my website (www.eduardlid.com/insert data)!

vibhaJ 126 Master Poster · Answer 5 · 2011-05-02T19:28:20+00:00

Can you try this code?

echo 'Query:'.$sql = "SELECT * FROM Products";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
	echo $row['Description'] . " " . $row['Price'] . " " . $row['Quantity'];
	echo "<br />";
}

eduardc 0 Light Poster · Answer 6 · 2011-05-02T20:42:00+00:00

Can you try this code?

echo 'Query:'.$sql = "SELECT * FROM Products";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
	echo $row['Description'] . " " . $row['Price'] . " " . $row['Quantity'];
	echo "<br />";
}

Thanks! However, neither are these improvements good:

Query:SELECT * FROM Products
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/eduardli/public_html/web_designer/insert.php on line 23

rpv_sen 59 Junior Poster · Answer 7 · 2011-05-02T21:25:28+00:00

Hi

Pls try this code

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

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("table_name", $con);


mysql_query ("INSERT INTO Products (Description, Price, Quantity) 
VALUES ('$description', '$price', '$quantity')");

$q="select * from Products";
$result=mysql_query($q,$con);
while($row = mysql_fetch_array($result))
{
echo $row['description'] . " " . $row['price'] . " " . $row['quantity'];
echo "<br />";
}

mysql_close($con);
?>

eduardc 0 Light Poster · Answer 8 · 2011-05-02T22:25:15+00:00

Hi

Pls try this code

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

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("table_name", $con);


mysql_query ("INSERT INTO Products (Description, Price, Quantity) 
VALUES ('$description', '$price', '$quantity')");

$q="select * from Products";
$result=mysql_query($q,$con);
while($row = mysql_fetch_array($result))
{
echo $row['description'] . " " . $row['price'] . " " . $row['quantity'];
echo "<br />";
}

mysql_close($con);
?>

Thanks!
But neither these improvements are good:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/eduardli/public_html/web_designer/insert.php on line 23

eduardc 0 Light Poster · Answer 9 · 2011-05-02T23:21:49+00:00

Thanks!
But neither these improvements are good:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/eduardli/public_html/web_designer/insert.php on line 23

I have checked the php file, that´s ok! Could the error be in my form.html?

kekkaishi 18 Junior Poster · Answer 10 · 2011-05-02T23:42:36+00:00

Hi,
"Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given"
This could mean that $result (the first parameter passed) is boolean. Just to check this, could you echo $result right before you start the while loop?

Also, I suppose you have made changes to your original code in insert.php. If you could perhaps post your current code...

TheSassyDragon 0 Newbie Poster · Answer 11 · 2011-05-03T02:04:52+00:00

Post again what code your using, it looks like your not initializing $result right, you need to be setting it to a SQL handle and the phrase "boolean given" looks like its either returning a false value or your not querying what you want to be.

Woops sorry missed the second page looks like you beat me to it.

eduardc 0 Light Poster · Answer 12 · 2011-05-03T08:35:25+00:00

Hi,
"Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given"
This could mean that $result (the first parameter passed) is boolean. Just to check this, could you echo $result right before you start the while loop?
Also, I suppose you have made changes to your original code in insert.php. If you could perhaps post your current code...

<?php
$con = mysql_connect("localhost","eduardli_user","-z.x,c");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("eduardli_company", $con);

mysql_query ("INSERT INTO Products (Description, Price, Quantity)
VALUES ('$description', '$price', '$quantity')");

$q="select * from Products";
$result = mysql_query($q,$con);
while($row = mysql_fetch_array($result))
{
echo $row . " " . $row . " " . $row;
echo "<br />";
}

mysql_close($con);
?>

</body>
</html

happytogether 6 Junior Poster in Training · Answer 13 · 2011-05-03T10:02:13+00:00

if you intend to search for specific result use where clause

//if you are trying to concat dont forget to add (.)
mysql_query("SELECT*FROM Products where field_name ='".$result."'");

karthik_ppts 81 Posting Pro · Answer 14 · 2011-05-03T10:06:59+00:00

Just echo your query as follows

echo $q="select * from Products";

In the output copy this query and paste it in SQL section of your PhpMyAdmin. See what is the result?

happytogether 6 Junior Poster in Training · Answer 15 · 2011-05-03T10:19:06+00:00

//youre code
mysql_query("SELECT*FROM Products='"$result"'");

//it should be
$string= 'test'; //youre specific string you want to search in you're database
$result = mysql_query("SELECT*FROM Products where field_name='".$string."'");

Kannan R 0 Newbie Poster · Answer 16 · 2011-05-03T16:34:03+00:00

$result is not assigned with any value
and use
mysql_query("SELECT * FROM Products='$result'");
it may work

rpv_sen 59 Junior Poster · Answer 17 · 2011-05-03T17:12:28+00:00

rpv_sen 59 Junior Poster

12 Years Ago

HI

Its Working fine for me.

Pls check the attachment

danitest.php (0.44 KB)

<html>
<head>
<style type="text/css">

#desc
{
width:10%;
height:5%;
}
</style>
<title>POST</title>
</head>

<body>


<form action='insert.php' method="post">

Description: <input type="text" name="description" id="description"><br />
Price: <input type="text" name="price" id="price" /><br />
Quantity: <input type="text" name="quantity" id="quantity" />
<input type='submit' value='submit' />


</form>



</body>
</html>

insert.php (0.59 KB)

<?php
$con = mysql_connect("localhost","root","123");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("test", $con);


mysql_query ("INSERT INTO Products (description, price, quantity) 
VALUES ('$description', '$price', '$quantity')");

$q="select * from Products";
$result=mysql_query($q,$con);
while($row = mysql_fetch_array($result))
{
echo $row['description'] . " " . $row['price'] . " " . $row['quantity'];
echo "<br />";
}

mysql_close($con);
?>

This attachment is potentially unsafe to open. It may be an executable that is capable of making changes to your file system, or it may require specific software to open. Use caution and only open this attachment if you are comfortable working with zip files.

products_sql.zip (0.64 KB)

eduardc 0 Light Poster · Answer 18 · 2011-05-04T04:36:46+00:00

//youre code
mysql_query("SELECT*FROM Products='"$result"'");

//it should be
$string= 'test'; //youre specific string you want to search in you're database
$result = mysql_query("SELECT*FROM Products where field_name='".$string."'");

Thanks! But also doesn´t work!

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/eduardli/public_html/web_designer/insert.php on line 24

eduardc 0 Light Poster · Answer 19 · 2011-05-04T05:40:12+00:00

HI
Its Working fine for me.
Pls check the attachment

Thanks! (I used your attachements, however it isn´t working?)

Table 'eduardli_company.Products' doesn't exist

eduardc 0 Light Poster · Answer 20 · 2011-05-04T06:16:46+00:00

Thanks! (I used your attachements, however it isn´t working?)
Table 'eduardli_company.Products' doesn't exist

It´s working now! However, I want the latest (only one!) output!

rpv_sen 59 Junior Poster · Answer 21 · 2011-05-04T10:12:11+00:00

hi

you have to use one more field as serial number, to fetch the latest one try to use the below code with your requirement

<?php
$latest="select * from tablename order by sno desc limit 0,1";
$res=mysql_query($latest,$conn);
$row=mysql_fetch_array($res);
?>

jack4321 -3 Newbie Poster · Answer 22 · 2011-05-04T11:13:50+00:00

Hi

Pls try this code

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

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("table_name", $con);


mysql_query ("INSERT INTO Products (Description, Price, Quantity) 
VALUES ('$description', '$price', '$quantity')");

$q="select * from Products";
$result=mysql_query($q,$con);
while($row = mysql_fetch_array($result))
{
echo $row['description'] . " " . $row['price'] . " " . $row['quantity'];
echo "<br />";
}

mysql_close($con);
?>

This Coding in perfect just check your html tags.

happytogether 6 Junior Poster in Training · Answer 23 · 2011-05-04T15:37:00+00:00

Thanks! But also doesn´t work!

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/eduardli/public_html/web_designer/insert.php on line 24

can you post youre new/latest code of insert into query?so we can see what is the error.

eduardc 0 Light Poster · Answer 24 · 2011-05-04T18:21:52+00:00

hi
you have to use one more field as serial number, to fetch the latest one try to use the below code with your requirement
<?php
$latest="select * from tablename order by sno desc limit 0,1";
$res=mysql_query($latest,$conn);
$row=mysql_fetch_array($res);
?>

Thanks!

eduardc 0 Light Poster · Answer 25 · 2011-05-05T04:44:16+00:00

hi
you have to use one more field as serial number, to fetch the latest one try to use the below code with your requirement
<?php
$latest="select * from tablename order by sno desc limit 0,1";


$res=mysql_query($latest,$conn);
$row=mysql_fetch_array($res);
?>

Hi,

I don´t understand your reply well!
I have to make an extra field to my table (Products) e. g. in phpMyAdmin?
And your code: Do I have to add it to mine or do I have to change mine with it?

eduardc 0 Light Poster · Answer 26 · 2011-05-05T06:09:48+00:00

hi
you have to use one more field as serial number, to fetch the latest one try to use the below code with your requirement
<?php
$latest="select * from tablename order by sno desc limit 0,1";
$res=mysql_query($latest,$conn);
$row=mysql_fetch_array($res);
?>

I changed the file, but got this error:

Parse error: syntax error, unexpected ')' in /home/eduardli/public_html/web_designer/insert.php on line 23

<?php
$con = mysql_connect("localhost","eduardli_user","-z.x,c");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("eduardli_company", $con) or die(mysql_error());

mysql_query ("INSERT INTO Products (description, price, quantity)
VALUES ('$description', '$price', '$quantity')") or die(mysql_error());

$latest="select * from Products order by serial no. desc limit 0,1";
$result = mysql_query($latest,$con) or die(mysql_error());
$row = mysql_fetch_array($result));
{
echo $row . " " . $row . " " . $row;
echo "<br />";
}

mysql_close($con);
?>

</body>
</html>

kekkaishi 18 Junior Poster · Answer 27 · 2011-05-05T06:33:32+00:00

Hi,

There is an extra parenthesis in the following line.

$row = mysql_fetch_array($result)); //the last parenthesis is not required
//it should be
$row = mysql_fetch_array($result);

Also, wrap your code in CODE tags to make your codes easily readable.

Hope this helps.

eduardc 0 Light Poster · Answer 28 · 2011-05-05T07:55:33+00:00

Hi,
There is an extra parenthesis in the following line.
$row = mysql_fetch_array($result)); //the last parenthesis is not required
//it should be
$row = mysql_fetch_array($result);
Also, wrap your code in CODE tags to make your codes easily readable.
Hope this helps.

I got this error message:

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 'desc limit 0,1' at line 1

<?php
$con = mysql_connect("localhost","eduardli_user","-z.x,c");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$description = $_POST["description"];
$price = $_POST["price"];
$quantity = $_POST["quantity"];

mysql_select_db("eduardli_company", $con) or die(mysql_error());

mysql_query ("INSERT INTO Products (description, price, quantity)
VALUES ('$description', '$price', '$quantity')") or die(mysql_error());

$latest="select * from Products order by serial no. desc limit 0,1";
$result = mysql_query($latest,$con) or die(mysql_error());
$row = mysql_fetch_array($result);
{
echo $row . " " . $row . " " . $row . " " . $row;
echo "<br />";
}

mysql_close($con);
?>

</body>
</html>

Php syntax error?

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