This is my MySQL statement, I want to search record by status or description.. this statement works fine in phpMyAdmin, but it is not working in php script.. Any Suggestions Please..
$result = mysqli_query($mysqli,
"SELECT * FROM `statuses`
where statuses.`status` LIKE '%$search%' OR
statuses.`description` LIKE '%$search%'");
I hope you write everything correct but there may be error in how you fetching data. Here are the things you need to check.
check your connection string
If you are retrieving data then use something like below
while($row = $result->fetch_array())
{
echo $row['example_col_name'];
}
You can do one more thing if everything alright store your query to a variable and echo out that one then you will see what query is passing .
Related
I have a multi query select which half works. The first query is straight forward.
$sql = "SELECT riskAudDate, riskClientId, RiskNewId FROM tblriskregister ORDER BY riskId DESC LIMIT 1;";
The second one doesn't seem to work even when I do it on its own:
$sql ="SELECT LAST(riskFacility) FROM tbleClients";
If I get rid of the LAST it returns the first entry in that field of the table. I want to use the LAST to get the LAST entry in that field.
When I do the first query on its own I get the data returned and I can echo it to the screen. When I add the second (with out the LAST) I get nothing. Here is what I am using
$result = $conn->query($sql);
if ($result == TRUE){
$r = $result->fetch_array(MYSQLI_ASSOC);
echo $r['riskAudDate'];
echo $r['riskClientId'];
echo $r['RiskNewId'];
echo $r['riskFacility'];
echo "<pre>";
print_r($r);
echo "</pre>";
}
The last bit is just for me to see whats in the array and just for testing.
So I have worked out that its the results array that is not right.
If I change the actual query to multi query I get this:
Call to a member function fetch_array() on boolean
So the array bit seems to be wrong for a multi query. The data returned is one row from each table. It works for the top query but add in the second (which I'm not sure is correct anyway) and the whole things crashes. So I guess it's a two part question. Whats wrong with my inserts and whats wrong with my returned array?
There is no last() function in mysql, it is only supported in ms access, if I'm not much mistaken. In mysql you can do what you do in the 1st query: do an order by and limit the results to 1.
According to the error message, the $conn->query($sql) returns a boolean value (probably true), therefore you cannot call $result->fetch_array(MYSQLI_ASSOC) on it. Since we have no idea what exactly you have in $sql variable, al I can say is that you need to debug your code to detrmine why $conn->query($sql) returns a boolean value.
Although it is not that clear from mysqli_query()'s documentation, but it only supports the execution of 1 query at a time. To execute multiple queries in one go, use mysqli_multi_query() (you can call this one in OO mode as well, see documentation). However, for security reasons I would rather call mysqli_query() twice separately. It is more difficult to execute a successful sql injection attack, if you cannot execute multiple queries.
It seems to me you are trying to do two SQL-queries at once.
That is not possible. Do a separate
$result = $conn->query($sql);
if ($result == TRUE){
while( $r = $result->fetch_array(MYSQLI_ASSOC)) {
...
}
}
for each SQL-query.
concerning :
$sql ="SELECT LAST(riskFacility) FROM tbleClients";
since the last function does not exists in MySQL i would recommend doing a sort like this(because i don't know what you mean with last )
$sql ="SELECT riskFacility FROM tbleClients order by riskFacility desc limit 0,1";
I am kind of stuck since I never tried to get something from the database in this way.
My problem:
I have a single line of code that goes to a website for example src = "www.youtube.com" however that link is stored in a database that has 1 single table called link so i retrieve this link like this
$sql = $db->prepare("SELECT link FROM youtubevid");
I have always used id's but for this table there is no need because it will get updated and i only need that 1 link. My question is how do i echo this?
Help is much appreciated
Thanks!
I guess you are using PDO to do that, if that's the case then you need to execute the query and fetch the result, after the line that you have specified i.e after the.
$sql = $db->prepare("SELECT link FROM youtubevid");
Here is the complete code:
// prepare the query
$sql = $db->prepare("SELECT link FROM youtubevid");
// execute it
$sql->execute();
// fetch the result
$result = $sql->fetch();
// echo out the column.
echo $result['link'];
I have a weird problem please take a look at this query:
select * from myfriend where name like "%n%";
when execute this query on phpMyAdmin the query returned correct results, but when execute it using php no result returned.
please note this query executed in drupal 6.
what is the problem with char "n" and PHP?
Percent signs are used as placeholders in Drupal 6 queries, so you need to escape them:
$query = db_query('select * from myfriend where name like "%%n%%"');
$searchChar = "n";
$query = "SELECT * FROM `myfriend` WHERE `name` LIKE '%" . $searchChar . "%'";
Then use the $query variable in your statement.
Eg:
$mysql->query($query);
mysql_query($query);
Your query is perfect. Give some brief on it. You can check if your connection of database from php to mysql is correct. You can echo that query from php file and run into phpmyadmin if that gives correct output then surely database connectivity problem will be there.
There is absolutely no issues with any character in php.
Im a php/mySQL newbie and am trying to get the hang of it. I have code to detect whether i get a username/password match, and now im trying to get the userid field so i can update the record. Heres what I have so far:
$sql = "SELECT username FROM users WHERE username='$username' AND password='$password'";
$result = $link->query($sql) or die(mysqli_error());
Using print_r($result) shows that there is an item, but im lost from here on out.
Try this.
$sql = "SELECT username FROM users WHERE username='$username' AND password='$password'";
$result = $link->query($sql) or die(mysqli_error());
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$userID= $row['username'] ;
// If you need other field as userID just change the sql and the index of $row according to that.
}
EDIT
If you want to get only one row.
if($result->num_rows==1)
{
$row = $result->fetch_array(MYSQLI_ASSOC);
$userID = $row["username"];
}
Perhaps this will help. In any programming language, running an SQL query is going to consist of these steps:
Build the text of the SQL statement that you want to run.
(Optional) If your statement involves the use of parameters (or "placeholders"), prepare an array of the parameter-values that are to be substituted for each of them.
("Prepare" and...) "Run" the query, on some previously-opened "database connection." (In your example, "$link" must correspond to that connection.) This gives you a handle (you called it "$result") that corresponds to the zero-or-more rows that were returned by that query.
Now, use that handle to retrieve each of these rows, one at a time, until there are no more or until you're tired of doing it.
(Optional) Be neat and tidy and "close" the handle, thus indicating to the database system that it can discard all of the resources it was using to furnish those rows to you.
"Those, in simple terms, are the basic steps that every program in the known universe are going to go through," and if you now browse again through the PHP documentation, you'll see that there are functions that correspond to each of these steps. Browse through the chapters you've been reading and see if you can now match the up to the scenario I just described. HTH...
This is really getting frustrating. I have a text file that I'm reading for a list of part numbers that goes into an array. I'm using the following foreach function to search a database for matching numbers.
$file = file('parts_array.txt');
foreach ($file as $newPart)
{
$sql = "SELECT products_sku FROM products WHERE products_sku='" . $newPart . "'";
$rs = mysql_query($sql);
$num_rows = mysql_num_rows($rs);
echo $num_rows;
echo "<br />";
}
The problem is I'm getting 0 rows returned from mysql_num_rows. I can type the sql statement without the variable and it works perfectly. I can even echo out the sql statement from this script, copy and paste the statement from the browser and it works. But, for some reason I'm not getting any records when I'm using the variable. I've used variables in sql statements tons of times, but this really has me stumped.
Try trimming and mysql_real_escape_string on your variable.
Check the source code of what is being echoed out and try to copy and paste that into PHPMyAdmin or something similar.
file includes newlines in the array elements. This may explain why it works when you copy the browser output but not in the script. You can try either:
$file = file('parts_array.txt', FILE_IGNORE_NEW_LINES);
or:
$sql = "SELECT products_sku FROM products WHERE products_sku='" . trim($newPart) . "'";
Note: Even though you're importing from a file of your own making, you can never be 100% sure that inject-able data hasn't been inserted into it. You should make sure to properly escape any data with mysql_real_escape_string. Even better would be using PDO prepared statements instead.
Obviously your code does something different than you expect. Running a successful query, for one: you don't check the return value of the mysql_query call, so you cannot be sure the query executed ok.
My idea:
dump your sql statement from the foreach
check the return code of the mysql_query
What does your parts_array.txt file look like? Do SKU numbers contain the ' character?
Can you please try this:
$file = file('parts_array.txt');
foreach ($file as $line_num => $line)
{
$sql = "SELECT products_sku FROM products WHERE products_sku='$line'";
echo $sql;
$rs = mysql_query($sql);
$num_rows = mysql_num_rows($rs);
echo $num_rows;
echo "<br />";
}
You might want to check for a mysql_error. It sounds like you've already verified the variable and have copied the query into a database interface like PHPMyAdmin or Query Browser, but if you haven't, I would recommend that.
After, verify that a very basic query will work, like SELECT * FROM Products. That will tell you if there is a problem outside of the query.
Overall, I would say the strategy would be to break the problem down into possible problem areas, like database, connection, query, errors, etc. Try to eliminate them one at a time until the problem is apparent. In other words, list the possibilities and cross them off one at a time.
I've encountered problems like this before; the trick is usually to start echoing things until you see the problem, and don't work off of assumptions.
I know this is pretty old now- but I'd like to help out others who may also be facing a similar problem with SQL statements that need to contain a potentially infinite number generated search parameters.
The code in the askers question is perfectly valid (for the avoidance of doubt) [see below]:
$file = file('parts_array.txt');
foreach ($file as $newPart)
{
$sql = "SELECT products_sku FROM products WHERE products_sku='" . $newPart . "'";
$rs = mysql_query($sql);
$num_rows = mysql_num_rows($rs);
echo $num_rows;
echo "<br />";
}
Their problem lies in the formatting of their text file ('parts_array.txt'). The root cause of the issue can be tracked down by dumping the information sent back by the server. Alternatively- they can try writing an SQL query in PHPMyAdmin and pasting in some or all of the data in their text file. MySQL will happily torment them until they find the problem.
For those trying to implement a variable based SQL query- the above is the way to go.
If you are trying to get data from an array, instead of a text file- you could do something like the following:
foreach ($array as $array_stuff)
{
$search_query = "SELECT * FROM table WHERE id='" . $array_stuff . "'";
$rs = mysqli_query($database_connection, $search_query);
$table_rows = mysqli_fetch_assoc($rs);
echo $table_rows['id']." - ".$table_rows['desc'];
echo "<br />";
}
/* free result set */
mysqli_free_result($rs);
This would output your data like this:
1001 - data 1 1002 - data 2 1003 - data 3
Note: The use of "mysql" functions are actively discouraged by MySQL. Therefore the second example I have given above is more up-to-date with current technologies, and using "mysqli" instead.
Also important
If you are here from a Google search as a result of trying to get data from a database, using a complex SQL query- you might have already tried to do something like the example below (or be considering it).
Do not attempt to write a variable based SQL query as per the example below. It won't work and will be incredibly frustrating.
Based on recent technological advancements- the second example I have given (using "mysqli") is the correct way (if there is one) to achieve this.
Bad example:
if ($search_result = mysqli_query($dbh1, "SELECT FROM sic_codes WHERE id = (".foreach ($_POST['SIC_Codes'] as $sic_codes) {echo "'".$sic_codes."' OR id = '',";})) {
/* fetch associative array */
while ($search_row = mysqli_fetch_assoc($search_result)) {
echo $row["id"]." - ".$row["desc"]."<br/>";
}