I wrote the following code to select text from database,but when i echo the output it giving output as Resource id #4
mysql_select_db("xxxxx", $link);
$q = "SELECT start_of FROM `qr_table` WHERE id_qr =1";
$result = mysql_query ($q, $link);
echo $result;
i am new to sql,forgive me if its a stupid questain
Thanks in advance
I suggest you to read at least Php documentation about Mysql query function.
you are echoing out the connection. you need to do something with the results like loop through them please check http://www.php.net/manual/en/function.mysql-query.php
You can't print out the result from MySQL directly. Try mysql_fetch_assoc(), which loads the value of each column into an associative array. if you have multiple rows returned, it will move to the next one each time it is called, and return false when there are no more.
this help for you
<?php
$link=mysql_pconnect("localhost","root","")or die("Not connected".mysql_error());
mysql_select_db("test");
$query="select * from qr_table";
$result=mysql_query($query,$link)or die("Query failed".mysql_error());
print "<center><table border=1>\n";
while($line=mysql_fetch_array($result,MYSQL_ASSOC)){
print "\t<tr>\n";
foreach($line as $col_value){
print "\t\t<td>$col_value</td>\n";
}
print "\t</tr>\n";
}
print "</table>\n</center>";
?>
Related
I'm writing a php file, and I want to show two tables by executing 2 seperate queries, and store them in $result and $result_bike. However, when I try to open the html page for this action form, it only shows the table for the first query, and gives the error " Commands out of sync; you can't run this command now" at the place of the second table.
Also, I don't want to combine these two tables, as they show entirely different information and I want to insert some text explaining each table.
I think there might have something to do with not able to print two tables for php (which I doubt)? What change should I make?
Thank you in advance for the help!
$result = mysql_query("CALL CrashTypeRate_Ped('$city')", $conn);
if (!$result){
echo "Fail to retrieve result for pedestrian crashes!\n";
print mysql_error();
} else {
echo "<table border=1>\n";
echo "<tr><td>CrashType</td><td>Count</td><td>TotalCount</td></tr>\n";
while ($myrow = mysql_fetch_array($result)) {
printf("<tr><td>%s</td><td>%s</td><td>%s</td></tr>\n", $myrow["crash_type"], $myrow["type_count"], $myrow["total_count"]);
}
echo "</table>\n";
}
$result_bike = mysql_query("CALL CrashTypeRate_Bike('$city')", $conn);
if (!$result_bike) {
echo "Fail to retrieve result for bike crashes!\n";
print mysql_error();
} else {
echo "got here!!!!!!";
echo "<table border=1>\n";
echo "<tr><td>CrashType</td><td>Count</td><td>TotalCount</td></tr>\n";
while ($myrow = mysql_fetch_array($result_bike)) {
printf("<tr><td>%s</td><td>%s</td><td>%s</td></tr>\n", $myrow["crash_type"], $myrow["type_count"], $myrow["total_count"]);
}
echo "</table>\n";
}
Here is from PHP Documentation user comments. Hope this helps
Link to PHP Documentation
When calling multiple stored procedures, you can run into the following error: "Commands out of sync; you can't run this command now".
This can happen even when using the close() function on the result object between calls.
To fix the problem, remember to call the next_result() function on the mysqli object after each stored procedure call. See example below:
<?php
// New Connection
$db = new mysqli('localhost','user','pass','database');
// Check for errors
if(mysqli_connect_errno()){
echo mysqli_connect_error();
}
// 1st Query
$result = $db->query("call getUsers()");
if($result){
// Cycle through results
while ($row = $result->fetch_object()){
$user_arr[] = $row;
}
// Free result set
$result->close();
$db->next_result();
}
// 2nd Query
$result = $db->query("call getGroups()");
if($result){
// Cycle through results
while ($row = $result->fetch_object()){
$group_arr[] = $row;
}
// Free result set
$result->close();
$db->next_result();
}
else echo($db->error);
// Close connection
$db->close();
?>
I've been on this for some days now. At first I thought the problem was in binding the parameters but I've simplified back to a basic mysqli page and still can't find the error. I'm passing the key for one of the rows in the search page before this onto this page so that I can show more details of the item which was selected.
I added an echo to test the the isset which prints correctly, also it puts the Key into the URL. If I leave out the WHERE Key = '$Key' it prints out the entire dataset. If I replace $row['Key'] with $Key it prints the whole dataset but with the selected key on every row.
This tells me that it is passing the key correctly and the print function is correct. I've tried using WHERE Key = $_GET['Key'] as well as $Key but neither work. I must be doing something basicly wrong here but after three days of trying every variation on the code I can think of, I have no more ideas.
<?php
$mysqli = new mysqli('localhost','user','password','database');
if ($mysqli->connect_error) {
die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
}
if(isset($_GET['Key'])){
$Key = $_GET['Key'];
echo "Got it";
}else{
echo "No input";
}
$results = $mysqli->query("SELECT * FROM engravers WHERE Key ='$Key'");
$img_url = "http://www.xxxxx.net/images/";
print '<table border="1" >';
while($row = $results->fetch_assoc()) {
print '<tr>';
print '<td>'.$row["Key"].'</td>';
print '<td>'.$row["Country"].'</td>';
print '<td>'.$row["Year"].'</td>';
print '<td>'.$row["Description"].'</td>';
print '<td>'.$row["Engraver1Surname"].'</td>';
print '<td>'.$row["Designer1Surname"].'</td>';
print '<td>'.$row["Printer"].'</td>';
print '<td>'.'<img src="'.$img_url.$row['Images'].'" />'.'</td>';
print '</tr>';
}
print '</table>';
$results->free();
$mysqli->close();
?>
</body>
There are many SQL column names you should avoid. Please read: http://technet.microsoft.com/en-us/library/ms189822.aspx
Same Reserved Keywords are in MySQL.
If you use one of those just cover it with ``
$results = $mysqli->query("SELECT * FROM `engravers` WHERE `Key`='$Key'");
In your query you are using column key which is not allowed because this is a reserved keyword.
Man, you're already using mysqli, why are you still interpolating variables in your query? This is the way to pass a string $Key to the parser
$results = $mysqli->prepare("SELECT * FROM engravers WHERE `Key`= ?");
$results->bind_param("s", $Key);
while($row = $results->fetch_assoc()) {
... stuff ...
}
If that doesn't work, I'm sure it's not because the prepared statement is the problem.
PD: key is a reserved word, so please note the fieldname is wrapped in backticks.
Hi I am trying to fetch data from a particular coloumn from all rows.
Eg Situation:
DB Data: id, fbid, name
$sql = 'SELECT id FROM table WHERE table.fbid IN (1234,5678,4321)';
$sql_run = mysql_query($sql);
$sql_fetch = mysql_fetch_assoc($sql_run);
print_r($sql_fetch);
This returns the data when I test it using Sequel PRO or PHPmyAdmin.
But when I print the array it only displays one value.
Can you help me with a solution or tell me where I'm going wrong?
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
while($row = mysqli_fetch_array($result))
{
echo $row['FirstName'] . " " . $row['LastName'];
echo "<br>";
}
mysqli_close($con);
?>
Before using a function, or - at least - when it does not what you expect - it's always a good idea to read the function description in the manual page.
PHP provides extremely easy access to its manual pages. All you need to type in the address bar is php.net/function name. It takes less time than typing whole question on Stack Overflow, yet you will get exactly the same answer. Think of efficiency.
You need to loop through each row
$sql_run = mysql_query($sql) or die(mysql_error());
while ($sql_fetch = mysql_fetch_assoc($sql_run)) {
print_r($sql_fetch);
}
Here is my code:
$campagin_id = $_SESSION['campagin_id_for_camp'];
$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;
$conn=mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());
mysql_select_db($dbconfig['db_name'],$conn);
$exec_query =mysql_query($query) or die(mysql_error());
$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
The Problem is that I am not getting anything in $row I cant get into the while loop, nothing shows up when I try to echo the value of $row, No error Nothing. Can you help me to find a problem in my code ?
Ps : The database is their. I have checked for the query for the corresponding value of $campagin_id. and also when i tried to echo $exec_query it echoed this : Resource id #8
PPS : The database have more than 7 record for each id so it doesn't matter if I call mysql_fetch_array($exec_query) more than once before going in to the while loop. and for the $campagin_id in the session their are many records present in the database.
You have written $row=mysql_fetch_array($exec_query) and then you are echoing something. and you are using the same in while.
Instead of:
$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
Use this (as per my knowledge you should not use $row=mysql_fetch_array() once you have used before while):
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
If the query returns Resource id #8 then that means it was successful - ie there were no errors. There were probably no rows returned by that query, so no rows in your table that match the given campagin_id.
You are also calling mysql_fetch_array() twice separately, you shouldn't do that because your while loop will skip the first row because calling this moves the pointer in the result set forward by one.
Also you can't echo an array as you are trying to, if you want to see the contents of an array use print_r() or var_dump().
I suggest adding some code to handle no rows found:
if($exec_query && mysql_num_rows($exec_query) > 0)
{
while ($row=mysql_fetch_array($exec_query)){
echo "Row: " . print_r($row, true);
}
}
else
{
echo 'None found';
}
Try this code.
<?
$campagin_id = $_SESSION['campagin_id_for_camp'];
$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;
mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());
mysql_select_db($dbconfig['db_name']);
$exec_query =mysql_query($query) or die(mysql_error());
while ($row=mysql_fetch_assoc($exec_query)) {
echo "<br/> row = <pre>".print_r($row)."</pre><br/>";
}
?>
In the PHP code below i want to select all the column values of a table and make them options of a select form. The result is that i dont get any options at all. Could someone help? Thanks
<?php
// ....
$userid=$_SESSION['userid'];
echo "<select>";
$sql = "SELECT * FROM users where userid='".$userid."'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
echo "<option>" .$row['company']. "</option>";
}
echo "</select>";
mysqli_close();
?>
It may just be a typo but you have mixed mysql and mysqli functions, you have stated mysql_query and mysql_fetch_array, whereas you have closed it with mysqli_close, could be an issue depending on what you wrote above this code or indeed if this is not just a typo.
Other things to try would be try the query in your mysql client / phpmyadmin and see if it comes up with any results or errors.