So what I need to do is find in my tables where id = id and name = name. I have tried this with no success.
$sql="SELECT * FROM $mysqltable WHERE id='$id' && name='$name'"
EDIT I found that I was missing a } bracket below this and the above code does work :/ Thank you everyone.
This is how i would have done it: Also try entering the table name instead of a variable.
$sql = "SELECT * From $mysqltable WHERE id='$id' AND name='$name'";
$result = mysql_query($sql) or die(mysql_error());
Related
while executing the below query i'm not getting sucess value
$name=qwe;
$result = mysql_query("SELECT *FROM USER WHERE name = $name");
Instead of above query if i put the below, I can able to get the appropriate ans:
$result = mysql_query("SELECT *FROM USER WHERE name = 'qwe'");
Can anyone give a solution for my first query??
Close the variable in quote and concatenate the string into query
$name= "qwe";
$result = mysql_query("SELECT *FROM USER WHERE name = '".$name."'");
one more note
please use mysqli for security purposes. Thanks
Use the following:
$result = mysql_query("SELECT *FROM USER WHERE name = '".$name."'");
P.S. I have not tested the above code but it should work.
select * from table_name where col_name = '".$variable."'
Directly use :
$name = 'qwe';
$result = mysql_query("SELECT * FROM USER WHERE name = '$name'");
And on more advise, you should never use mysql_* functions now as they have been deprecated in newer versions of PHP and will produce the notices.
$results = mysql_query("select * from doctorlist where assignednumber = '1231231234' ");
I need to change the number 1231231234 to a variable. If I change it to the code below it does not work. I have displayed the variable on the page so I know it is set.
$results = mysql_query("select * from doctorlist where assignednumber = '$phoneNumber' ");
Could someone please help. I know it is a small issue, but have been unable to fix it.
Perhaps split it like this
$sql_query = "select * from doctorlist where assignednumber='$phoneNumber'";
$results = mysql_query($sql_query);
or
$sql_query = "select * from doctorlist where assignednumber='".$phoneNumber."' ";
$results = mysql_query($sql_query);
First check your variable type with var_dump($phoneNumber) than do the following:
$results = mysql_query("select * from doctorlist where assignednumber = '".$phoneNumber."' ");
to improve readability and last if you expect an Integer cast your variable like:
(int)$phoneNumber
or if string do
mysql_real_escape_string($phoneNumber)
Try using the variable inside the query like this:
'{$phoneNumber}'
I am trying to create a php script that selects one row from my table by using the WHERE clause. The problem is the mysql query returns no rows. I know the variable is correct (its user submitted).
$title = mysql_real_escape_string($_REQUEST["title"]);
$query = mysql_query("SELECT * FROM links WHERE title ='$title'", $con)
or die ("Error: " . mysql_error());
I'm looking for any ideas that could fix my problem. I know the mysql is working properly because other queries execute fine. The title variable is correct; it is passed from a mysql on another page.
ps - I posted a similar question earlier, but worded it poorly and got results that didn't address the problem
Try this:
$query = mysql_query("SELECT * FROM links WHERE title ='$title' limit 1")
or die ("Error: " . mysql_error());
Sorry, I'm new here and I couldn't find the button to comment on the original question.
But you mentioned the request was user submitted. Are they typing it or is it a selection like from a select box or radio button? I'm asking because does the requested title even exist in the DB?
Anyway, what is your result if you use the following?:
$query = mysql_query("SELECT * FROM links WHERE title LIKE '%".$title."%'")
or die ("Error: " . mysql_error());
If not, then there's definitely no match in the DB.
try this query
mysql_query("SELECT * FROM links WHERE title like '$title%'", $con)
Try this query:
$query = mysql_query("SELECT * FROM links WHERE title LIKE '%{$title}%'");
or maybe this to check formatting:
$sql = sprintf("SELECT * FROM links WHERE title LIKE '%%%s%%'", $title);
$query = mysql_query($sql);
In my case I had empty space before and after the variable name, like this
$query = "select * from user where user_name = ' $user_name ' ";
and this results in comparing userName with [empty_space userName empty_space ] which doesn't exist in the database.
your query should be like this
$query = "select * from user where user_name = '$user_name' ";
I'm having troubles with a PHP code. The problems come when I execute a SQL query with a PHP variable inside it. The result doesn't shows anything, and the field in the database is not empty. Tried out with a static id (not variable) and the query works fine.
Do you know where I'm doing it wrong?
The query code:
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = `'.$videoSeleccionado.'`;", $conexion);
Try this:
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = `'.$videoSeleccionado.'`;", $conexion) or die(mysql_error());
That will give you an error message.
The problem is that you use both ` and ' as escape characters as the same time.
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = '$videoSeleccionado';", $conexion);
should work.
Often things will be more clear when you echo the query so you can see the final result.
you are using double quotes on your query so there is no need for the dot . operator
$consultaVideoSeleccionado1 = mysql_query("SELECT * FROM videos WHERE idvideo = '$videoSeleccionado'", $conexion);
You are connecting string wrong! You are using ' single quote, but you should use double/none.
Try this query:
mysql_query("SELECT * FROM `videos` WHERE `idvideo` = '$videoSeleccionado';", $conexion)
Or
mysql_query("SELECT * FROM `videos` WHERE `idvideo` = '{$videoSeleccionado}';", $conexion)
Or
$q = "SELECT * FROM `videos` WHERE `idvideo` = '%s';";
mysql_query(sprintf($q, $videoSeleccionado), $conexion)
Edit:
If it's still not working problem can be in query, try checking if it is using mysql_error()(1) or try dumping query(2).
Example(1):
mysql_query("SELECT * FROM `videos` WHERE `idvideo` = '".$videoSeleccionado."';", $conexion) or die(mysql_error());
Example(2):
$q = "SELECT * FROM videos WHERE idvideo = '$videoSeleccionado';";
var_dump($q);
mysql_query($q, $conexion)
I need to read a text file, query the database table with that name, and store that table's data in another table. So far I have written this code but I don't know why it's not working.
foreach ($lindb as $namedb) {
$query = "SELECT * FROM ntable WHERE name =" .$namedb. "";
$result = mysql_query($query);
while ($r = mysql_fetch_array($result)) {
$query = "INSERT INTO ndtable (name,details,address,login,country) VALUES (\"".$r["name"]."\", \"".$r["details"]."\", \"".$r["address"]."\", \"".$r["login"]."\", \"".$r["country"]."\")";
mysql_query($query);
}
}
You don't have quotes around $namedb
ie. SELECT * FROM ntable WHERE name =" .$namedb. ""; should be SELECT * FROM ntable WHERE name ='" .$namedb. "'";
I suggest a SELECT INTO would be the better choice... and please post the error so we are able to help...