0

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>

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

15
Contributors
85
Replies
88
Views
6 Years
Discussion Span
Last Post by karthik_ppts
0

Hi

In line number 21 the statement is

mysql_query("SELECT*FROM Products='"$result"'");
Please change as mysql_query("SELECT*FROM Products='$result'");
0

you need to replace this line.

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

to

$result = mysql_query("SELECT * FROM Products");
0

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)!

0

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 />";
}
0

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

0

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);
?>
0

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

0

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?

0

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...

0

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.

Edited by TheSassyDragon: n/a

0

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...

<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')");

$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

0

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."'");
0

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?

0
//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."'");
1

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.

Votes + Comments
useful post
0

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

0

HI

Its Working fine for me.

Pls check the attachment

Attachments
<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
$con = mysql_connect("localhost","root","123");
if (!$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($con);
?>
0
//youre code
mysql_query("SELECT*FROM Products='"$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

0

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

0

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!

0

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);
?>
0

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.

0

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.

0

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!

0

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?

0

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

<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) 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>

0

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.

0

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

<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) 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>

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