i want to search data in an array .How can i search using mysql select command. i wrote the query as
$this->db->query("SELECT * FROM client_details WHERE dob IN ({implode(',', $data})");
$data is an array of dates .Please Help me for solving this..
try with
$data =array('0'=>'2015-01-23','1'=>'2015-01-22','2'=>'2015-01-21');
$tmp = implode('","', $data);
$tmp ='"'.$tmp.'"';
$sql= 'SELECT * FROM client_details WHERE dob IN ('.$tmp.')';
echo $sql;
$this->db->query($sql);
You can do something like this:
Option 1
$tmp = implode(',', $data);
$this->db->query("SELECT * FROM client_details WHERE dob IN ($tmp)");
Option 2
$this->db->query("SELECT * FROM client_details WHERE dob IN (".implode(',', $data).")");
Also, there is a mistake in your bracket formation.
It should be dob IN ({implode(',', $data)}"); [Curly braces should close after the closing parenthesis].
Related
I am running problems in implementing LIKE in PDO
I have this query:
$query = "SELECT * FROM tbl WHERE address LIKE '%?%' OR address LIKE '%?%'";
$params = array($var1, $var2);
$stmt = $handle->prepare($query);
$stmt->execute($params);
I checked the $var1 and $var2 they contain both the words I want to search, my PDO is working fine since some of my queries SELECT INSERT they work, it's just that I am not familiar in LIKE here in PDO.
The result is none returned. Do my $query is syntactically correct?
You have to include the % signs in the $params, not in the query:
$query = "SELECT * FROM tbl WHERE address LIKE ? OR address LIKE ?";
$params = array("%$var1%", "%$var2%");
$stmt = $handle->prepare($query);
$stmt->execute($params);
If you'd look at the generated query in your previous code, you'd see something like SELECT * FROM tbl WHERE address LIKE '%"foo"%' OR address LIKE '%"bar"%', because the prepared statement is quoting your values inside of an already quoted string.
Simply use the following:
$query = "SELECT * FROM tbl WHERE address LIKE CONCAT('%', :var1, '%')
OR address LIKE CONCAT('%', :var2, '%')";
$ar_val = array(':var1'=>$var1, ':var2'=>$var2);
if($sqlprep->execute($ar_val)) { ... }
No, you don't need to quote prepare placeholders. Also, include the % marks inside of your variables.
LIKE ?
And in the variable: %string%
$query = "SELECT * FROM tbl WHERE address LIKE ? OR address LIKE ?";
$params = array("%$var1%", "%$var2%");
$stmt = $handle->prepare($query);
$stmt->execute($params);
You can see below example
$title = 'PHP%';
$author = 'Bobi%';
// query
$sql = "SELECT * FROM books WHERE title like ? AND author like ? ";
$q = $conn->prepare($sql);
$q->execute(array($title,$author));
Hope it will work.
I want to get all the values in the database where date is between two given parameters. Following is my code. Here Date is $date1' and '$date2 are the given parameters.
$result1 = mysql_query("SELECT * FROM model WHERE Date BETWEEN('$date1' AND '$date2')");
while($row=mysql_fetch_array($result1)){
$d=$row['Date'];
echo $d;
}
You have an error in your syntax. Remove the parentheses after between:
$result1 = mysql_query("SELECT * FROM model WHERE Date BETWEEN '$date1' AND '$date2' ");
I'm guessing this is your question.
So I am trying to have a single script that accesses all values of a table based on a particular id. The script returns the values in an array using PHP:
Example:
// Select data from DB
$query = 'SELECT * FROM experiences WHERE user_id = ' . $id;
$result = mysqli_query($link, $query) or die (mysqli_error($link));
// Not sure this actually creates a 2D array...
$array = mysqli_fetch_array($result);
However, I realize that I need modified results for particular tasks, such as the row where a particular value is the highest.
How would I go about doing this, and does $array actually hold all the rows and respective fields?
TRY
'SELECT * FROM experiences WHERE user_id ='.$id.' HAVING MAX(column_name)
OR
"SELECT max(column_name), other column...
FROM experiences
WHERE user_id =".(int)$id
Assuming you have cleaned $id properly you can do this
// Select data from DB
$query = 'SELECT * FROM experiences WHERE user_id = ' . $id;
$result = mysqli_query($link, $query) or die (mysqli_error($link));
$data = array();
while($row = mysqli_fetch_array($result)) {
$data[] = $row;
}
and $data will contain the 2D array you're looking for.
If you want the row where a particular value is highest I suggest either looking into ORDER BY or MAX() depending on your needs.
I have the following query, and would like to list only the first match.
$first = $_GET['category'];
$first = $first[0] . "%";
$query = mysql_query("SELECT * FROM lyrics WHERE authorclean LIKE '".$first."'") or die(mysql_error());
(?category=b)
So DISTINCT could do this right? This is what I tried, but did not work:
$query = mysql_query("SELECT DISTINCT authorclean FROM lyrics WHERE authorclean LIKE '".$first."'") or die(mysql_error());
EDIT: Here is the full code:
function getCategory() {
$first = $_GET['category'];
$first = $first[0] . "%";
$query = mysql_query("SELECT DISTINCT authorclean FROM lyrics WHERE authorclean LIKE 'B%'") or die(mysql_error());
//$query = mysql_query("SELECT * FROM lyrics WHERE authorclean LIKE '".$first."'") or die(mysql_error());
if(mysql_num_rows($query) == 0) {
echo "Geen resultaten gevonden.";
} else {
while ($row = mysql_fetch_assoc($query)) { ?>
<p><?= $row['author']; ?></p>
<?php }
}
}
(B% is just for testing)
If I run this following query in the database directly I get two results. If I run with the code above I just get an empty page (except for the html thats already there).
SELECT DISTINCT authorclean FROM lyrics WHERE authorclean LIKE 'B%'
You should use LIMIT 1 to list only the first match.
If you have a a table "tbl_lyrics" with fields: author lyrics year and is filled for example as follows:
author_A lyrics_A year_A
author_A lyrics_A1 year_A1
author_A1 lyrics_A2 year_A
author_B lyrics_B1 year_B1
if you do
select distinct(author) from tbl_lyrics where author like '%author_A%'
you are going to get: author_A and author_A1. NOT the first one that matches.
If you want the first one that matches you can do:
select author from (select distinct(author) as author from tbl_lyrics where author like '%author_A%') where rownum <2;
this will return author_A only.
Limit is used with MySql but would not work with oracle databases
What's the best way to check whether the value is in the database?
Am I doing it correct?
$result = mysql_query("SELECT COUNT(*) FROM table WHERE name = 'John'");
$count = count($result);
you could use straight forward ,
mysql_num_rows() ;
eg :
$con = mysql_connect($host,$uname,$passwd)
mysql_select_db($dbase,$con);
$result = mysql_query($query,$con);// query : SELECT * FROM table WHERE name='jhon';
if( ! mysql_num_rows($result)) {
echo " Sorry no such value ";
}
Yes you are doing it right, if you are only concerned with checking if there are any records where name='john'
SELECT COUNT(*) FROM table WHERE name = 'John'
will return the no. of records where name field is 'John'. if there are no records then it will return 0, and if there are any records it will return the number of records.
But the above query will miss the entries where name is 'John Abraham' or 'V john', to include even these
you can modify your query like this.
SELECT COUNT(*) FROM table WHERE name like '%John%'
I'd say yes.
$result = mysql_query("SELECT COUNT(*) AS 'nb' FROM table WHERE name = 'John'");
$line = mysql_fetch_array($result, MYSQL_ASSOC);
$count = $line['nb'];
Will give you the number of matching rows.
$result = mysql_query("SELECT COUNT(*) as user FROM table WHERE name = 'John'");
$line = mysql_fetch_array($result, MYSQL_ASSOC);
$count = $line['user'];
if($count!=0)
{
echo "user exists";
}
else
{
echo "There is no such user";
}