I am trying to update a table (products) from another table (table 1). The model number, however, has many symbols in it, including the '/'. The data exporting works quite fine, except the case when there is '/'. It converts it to a date instead.
E.g the following cases
11/08/1944 should be 11/08/44
11/07/1944 should be 11/07/44
The following is what I have:
$query = mysql_query("SELECT `COL 1` AS id, `COL 7` AS model FROM `table 1` ORDER BY id ASC", $c2) or die(mysql_error());
while($r = mysql_fetch_array($query)):
$id = $r['id'];
$model = $r['model'];
mysql_query("UPDATE products SET model = '".mysql_real_escape_string($model)."' WHERE id = $id", $c1);
echo "MODEL: $model AND ID: $id <br />";
endwhile;
try those :
echo $model if its right .
look your column names if they are messed up.
Related
I have done this type of SELECT many times, but this time I can't get it to work. Any ideas, please?
$Name = "Dick";
$conn = mysqli_connect($server, $dbname, $dbpw, $dbuser);
$sql = "SELECT id FROM table WHERE $Name = table.first_name";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
$customer_id = $row['id'];
Database::disconnect();
echo "customer id = " . $customer_id;
If you really DO have a table named table it would be more appropriate to use back ticks around the name since the word TABLE is a reserved word in MySQL. You should also use single quotes around your variable if it contains a string:
$sql = "SELECT `id` FROM `table` WHERE `first_name` = '$Name'";
Other possible reasons if the query still doesn't work for you:
Make sure you have the connection parameters in the right order. It should be: mysqli_connect($server, $dbuser, $dbpw, $dbname).
You should be using fetch_array() instead of fetch_assoc() if you expect a one row result.
You are mixing PROCEDURAL STYLE with Object Oriented Style when using mysqli_connect() instead of mysqli(), at the same time using $result-> which is object oriented style. You should decide one style and stick with it.
This would be the procedural style of your query:
$Name = "Dick";
$conn = mysqli_connect($server, $dbuser, $dbpw, $dbname); // NOTE THE CHANGED ORDER OF CONNECTION PARAMETERS!
$sql = "SELECT `id` FROM `table` WHERE `first_name` = '$Name'";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);
$customer_id = $row['id']; // YOUR CUSTOMER ID
mysqli_free_result($result); // FREE RESULT SET
mysqli_close($conn); // CLOSE CONNECTION
And this would be the object oriented style:
$Name = "Dick";
$conn = new mysqli($server, $dbuser, $dbpw, $dbname);
$sql = "SELECT `id` FROM `table` WHERE `first_name` = '$Name'";
$result = $conn->query($sql);
$row = $result->fetch_array(MYSQLI_ASSOC);
$customer_id = $row['id']; // YOUR CUSTOMER ID
$result->free(); // FREE RESULT SET
$conn->close(); // CLOSE CONNECTION
I would recommend naming your table something else than table since it's a reserved word and could get you into parsing problems. The same goes with field names. More reading: https://dev.mysql.com/doc/refman/5.5/en/keywords.html
More about mysqli_fetch_array() and differences in procedural style and object oriented style use: http://php.net/manual/en/mysqli-result.fetch-array.php
$sql = "SELECT id FROM table WHERE '$Name' = table.first_name";
You simply need to concat the variable like this:
$sql = "SELECT id FROM table WHERE " . $Name . " = table.first_name";
I have a couple of MySQL tables where I run a query on like this:
$sql = "
SELECT my_item
FROM t1
, t2
WHERE t1.id='$id'
AND t2.spec IN (208, 606, 645)
AND t1.spec = t2.spec
";
Note I am using the WHERE IN.
Next I run a query and use WHILE to try to get the results:
$retval = mysql_query($sql) or die('Query failed: ' . mysql_error());
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
$myitem = $row['my_item'];
echo "My item is $myitem<br />\n";
}
This prints three results, each with a different value for $myitem, based on the three options from the IN clause in the SELECT statement at the beginning.
How can I extract and store each of these three values in a separate variable each?
Thank you!
use join
SELECT my_item FROM t1 join t2
on t1.spec=t2.spec
WHERE t1.id='$id'
AND t2.spec IN (208, 606, 645)
or create array
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
$myitem[] = $row['my_item'];
}
use this array likt that :-
foreach( $myitem as $myitems){
echo "My item is $myitems<br />\n";
}
or use indexing
echo "My item is $myitems[0]";
You can store the $row['my_item'] in an array.
Refer PHP arrays
I am trying to work with a voting database system. I have a form to display all the candidates per candidate type. I'm still trying to to explore that. But this time, I want to try one candidate type, lets say for Chairperson, I want to display all candidate names for that type in the ballot form. However, there's an error with the line where i declare $query and made query statements, can somebody know what it is. I am very certain that my syntax is correct.
function returnAllFromTable($table) {
include 'connect.php';
$data = array ();
$query = 'SELECT * FROM ' . $table. 'WHERE candidateId=1'; //ERROR
$mysql_query = mysql_query ( $query, $conn );
if (! $mysql_query) {
die ( 'Go Back<br>Unable to retrieve data from table ' . $table );
} else {
while ( $row = mysql_fetch_array ( $mysql_query ) ) {
$data [] = $row;
}
}
return $data;
}
As #Darwin von Corax says, I sure that you have a problem between $table and WHERE
Your query:
$query = 'SELECT * FROM ' . $table. 'WHERE candidateId=1';
If $table = 'Chairperson';
You have:
'SELECT * FROM ChairpersonWHERE candidateId=1';
The your query should be:
$query = 'SELECT * FROM ' . $table. ' WHERE candidateId=1';
I'm looking for a way to search a MySQL database for specific values and put these in an array.
The contacts table has name,email,group,Phone...
I would like to search the database by group and return the email adresses in an array, separated by , (comma) to use further in my code.
What is the best way to do this?
$result = mysqli_query($link,"SELECT * FROM Contacts WHERE Group='Group 1'")
or die(mysqli_error());
...
while($row = mysqli_fetch_array( $result ))
{
array ( row->email,...)
}
Here you have different options beginning with:
Update your select query to only display the email, if you don't require the other fields this is the way to go
In your loop just add only the email field to the array
$dataArray[] = $row->email;
$result = mysqli_query($link,"SELECT email FROM Contacts WHERE Group='Group 1'")
or die(mysqli_error());
$emailArray = [];
while($row = mysqli_fetch_array( $result ))
{
array_push($emailArray,$row->email);
}
$responseEmail = implode(",", $emailArray);
Hope this will works!
I found the way of what I was trying to do. I needed a selectbox with values from the database.
....
$query = mysqli_query($link,"SELECT * FROM XXX WHERE category='$category' AND stelplaats='$stelplaats'");
echo '<td><select name ="collega" required>';
echo '<option value"">---</option>';
while ($row2 = mysqli_fetch_array( $query ))
{
echo '<option value =" '.$row2['id'].'"';
echo '>'.$row2['name'].'</option>';
}
echo '</select></td></tr>';
....
I want to display four (4) items'name from these id:
Can I do like this?
SELECT item_name from items WHERE item_id IN ('001', '012', '103', '500')
or
SELECT item_name from items WHERE item_id = '001' or item_id = '012' or item_id = '103' or item_id = '500'
IN RESPONSE TO ALL ANSWERS
Well, most of the answers said it works, but it does not really work. Here is my code:
$query = "SELECT `item_name` from items WHERE item_id IN('s001','a012','t103','p500')";
$result = mysql_query($query, $conn) or die (mysql_error());
$fetch = mysql_fetch_assoc($result) or die (mysql_error());
$itemsCollected = $fetch['item_name'];
echo $itemsCollected;
The item_id is alphanumeric.
You can do either one, but the IN query is much more efficient for this purpose for any large queries. I did some simple testing long ago that revealed it's about 10 times faster to use the IN construct for this. If you're asking if the syntax is correct then yes, it looks fine, other than missing semi-colons to complete the statement.
EDIT: It looks like the actual question you were asking was "why do these queries only return one value". Well, looking at the sample code you posted, the problem is here:
$fetch = mysql_fetch_assoc($result) or die (mysql_error());
$itemsCollected = $fetch['item_name'];
echo $itemsCollected;
You need to loop through and iterate until there are no more results to be fetched, as Pax pointed out. See the PHP manual page for mysql_fetch_assoc:
$sql = "SELECT item_name from items WHERE item_id IN('s001','a012')";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
// Note: If you put extract($row); inside the following loop, you'll
// then create $userid, $fullname, and $userstatus
while ($row = mysql_fetch_assoc($result)) {
echo $row["userid"];
echo $row["fullname"];
echo $row["userstatus"];
}
mysql_free_result($result);
Yes, both those should work fine. What's the actual problem you're seeing?
If, as you say, only one record is being returned, try:
select item_name from items order by item_id
and check the full output to ensure you have entries for 001, 012, 103 and 500.
If both those queries only return one row, I would suspect not.
If they all do exist, check the table definitions, it may be that the column is CHAR(4) and contains spaces for the others. You may have genuinely found a bug in MySQL but I doubt it.
After EDIT:
This is a perl/mysql problem, not an SQL one: mysql_fetch_array() returns only one row of the dataset at a time and advances a pointer to the next.
You need to do something like:
$query = "SELECT item_name from items WHERE item_id IN('s001','a012')";
$result = mysql_query($query, $conn) or die (mysql_error());
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($result)) {
echo $row["item_name"];
}
Your ID field must be set to auto increment, i guess. i had problems with that once and i changed the auto increment to int. in the IN field if you pass the parameters to match against the auto increment variable you get back only the first parameter, the remaining generates an error.
Use mysql_fetch_assoc to get the query and assign the values from mysql_fetch_assoc query into a an array. Simple as that
$i=0;
$fullArray = array();
$query = mysql_query("SELECT name FROM users WHERE id='111' OR id='112' OR id='113' ")or die(mysql_error());
while($row = mysql_fetch_assoc($query)){
foreach ($row as $value) {
$fullArray[$i] = $value;
}
$i++;
}
var_dump($fullArray);
echo $fullArray[0]."<br/>".$fullArray[1]."<br/>".$fullArray[2];`
You could also use the mysql_num_rows function to tell you how many rows your query retrieved and then use that result to increment a for loop. An example.
$num_rows=mysql_num_rows($query_results);
for ($i=0; $i <$num_rows ; $i++) {
$query_array[]=mysql_fetch_assoc($query_results);
}