I output with the json_encode method. It does not show Turkish characters in the output I have received. What should I do to show Turkish characters? I am using mysql. Collation of the table = utf8_turkish_ci
<?php
$con=mysqli_connect(...);
header('Content-Type: text/html; charset=utf-8');
mysql_set_charset('utf8', $con);
$formID = $_POST['formID'];
$query = "SELECT ADI_SOYADI ...";
$result = mysqli_query($con,$query);
$rows = array();
while($r = mysqli_fetch_array($result)) {
$apptADI = $r['ADI_SOYADI'];
$results = Array("ADI_SOYADI" => $apptADI);
}
echo json_encode($results);
mysqli_close($con);
?>
It seems that prior to use json_encode()you should use utf8_encode()
https://www.geeksforgeeks.org/php-utf8_encode-function/
echo json_encode(utf8_encode($results));
There is also the parameter JSON_UNESCAPED_UNICODE that you can add to json_encode()
Related
I have this in my PHP.
$sql="select * from defaulttime";
$result = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($result)){
$row_set[] = $row;
}
echo trim(json_encode($row_set));
In my SQL some columns are Integer. but in JSON it shows as String.
can someone please tell me what I am doing wrong?
when I run the PHP, I get the below results. (Note Hour and Minute should be Integers)
[{"name":"Test","Hour": "6" ,"Minute":"45"}],
I want it to show string as string and Int as Int.
{"name":"Test","Hour": 6 ,"Minute":45}
You have to use JSON_NUMERIC_CHECK option in order to get numerics value,
look here: https://lornajane.net/posts/2011/php-returning-numeric-values-in-json
echo trim(json_encode($row_set,JSON_NUMERIC_CHECK));
Thanks all,
I got it working by using this code:
$result = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($result)){
$row_set[] = $row;
}
$json=json_encode($row_set,JSON_NUMERIC_CHECK);
echo $json;
I'm having issues with working with mysql's FIND_IN_SET and pdo. This is my code:
$statement = $conn->prepare("SELECT * FROM `artistInfo` WHERE FIND_IN_SET(':array', artistServices)");
$statement->execute(array(':array' => '2'));
while($row = $statement->fetch()){
echo $row['id'];
echo "<br />";
}
This doesn't produce any results. Am I doing something wrong? Thanks!
$statement = $conn->prepare("SELECT * FROM `artistInfo` WHERE FIND_IN_SET(':array', artistServices)");
$statement->execute(array(:array => '2'));
while($row = $statement->fetch()){
echo $row['id'];
echo "<br />";
}
The single quotes surrounding :array is why I wasn't getting any results
I keep getting an error using MySQL 5.5.27 when trying to concatenate some values. I've searched and seen a bunch of charset answers (which admittedly is a TAD over my head), but I've converted all my tables to Charset utf8-unicode-ci and still get the error.
Surely there is a way to concatenate these values, but I just don't know how. I'm an Oracle guy that is relatively new to MySQL.
Here is the SQL line:
concat(pl.last_name,'-',format(money,0))
I get:
#1270 - Illegal mix of collations (latin1_swedish_ci,IMPLICIT), (utf8_unicode_ci,COERCIBLE), (utf8_unicode_ci,COERCIBLE) for operation 'concat'
Any ideas?
If money is indeed a number inside a VARCHAR you could use cast.
Try this:
concat_ws(pl.last_name,'-',cast(money AS unsigned)); // This is with decimals.
concat(`pl.last_name,'-',substring_index(money,',',1)) // Without decimals. If you use . i.e. the American currency notation you can substitute , with an .
Edit
Your should first try: concat(pl.last_name,'-',format(money,0));
This a very basic php code you could use.
<?php
function selecting_data(){
$host = "host";
$user = "username";
$password = "password";
$database = "database";
$charset = "utf8";
$link = mysqli_connect($host, $user, $password, $database);
mysqli_set_charset($charset, $link);
IF (!$link) {
echo('Unable to connect to the database!');
} ELSE {
$query = "SELECT lastname, format(money,0) FROM mytable"; //Select query
$result = mysqli_query($link, $query);
while ($rows = mysqli_fetch_array($result, MYSQLI_BOTH)){
echo $rows['lastname']."<br>".$rows['money'] ;
}
}
mysqli_close($link);
}
?>
<html>
<head><title>title</title></head>
<body>
<?PHP echo selecting_data(); ?>
</body>
</html>
I am trying to get MySQL database into an xml file; here is my code:
<?php
header("Content-type: text/xml");
include 'dbc.php';
$query = "SELECT * FROM airports LIMIT 50";
$result = mysql_query($query, $link)
or die('Error querying database.');
$xml = new SimpleXMLElement('<xml/>');
while($row = mysql_fetch_assoc($result)) {
$draw = $xml->addChild('draw');
$draw->addChild('ident',htmlentities(iconv("UTF-8", "ISO-8859-1//IGNORE",$row['ident'])));
$draw->addChild('name',htmlentities(iconv("UTF-8", "ISO-8859-1//IGNORE",$row['name'])));
}
mysql_close($link);
$fp = fopen("links2.xml","wb");
fwrite($fp,$xml->asXML());
fclose($fp);
Here is the error Im getting:
XML Parsing Error: no element found
Location: /sql2xml2.php
Line Number 1, Column 2:
-^
What am I doing wrong???
Your XML is considered invalid in your XML reader because of the thrown warning, thus the XML Parsing Error: junk after document element issue.
As for the warning itself, you need to escape special entities (namely &, < and > in your content when adding it like that (using str_replace usually works well for only those 3 when it comes to XML, htmlentities may yield undesired effects, unless you supply PHP 5.4's ENT_XML1 mode).
Refer to a related answer for more information of why this happens.
If you want just to export MySQL database to local XML file you can use mysqldump tool:
mysqldump --xml -u username -p databasename [tablename] > filename.xml
Got it to work with this code:
<?
header("content-type:text/xml");
function getXML($query="SELECT * FROM airports limit 50")
{
include 'dbc.php';
$result = mysql_query($query, $link)
or die('Error querying database.');
$columns="";
echo "<xml>\n";
while($row=mysql_fetch_assoc($result))
{
$columns.="\t<airport>\n";
foreach($row as $key => $value)
{
$value = htmlentities(iconv("UTF-8", "ISO-8859-1//TRANSLIT",$value));
$value = htmlentities(iconv("UTF-8", "ISO-8859-1//IGNORE",$value));
$columns.="\t\t<$key>$value</$key>\n";
}
$columns.="\t</airport>\n";
}
echo $columns;
echo "</xml>\n";
}
getXML();
?>
basically I am try to get the php file to create another php file which has XHTML, PHP and MYSQL within it. I am trying to include a MYSQL query but the problem is that the apostrophe (') that I use to start the fwrite string get confused when it reaches a MYSQL query because it too also has an apostrophe but I just want that string to continue not end. What should I do for this to work? Thanks in advance for any help.
The code (just the relevant bit):
$filename = "websites/".$firstpage.".php";
$filehandle = fopen($filename, 'w') or die("error opening file");
$datawrite = '
$query = "SELECT websitetitle FROM $weblist WHERE webID='$webID'";
$result = mysql_query ($query);
$row = mysql_fetch_array( $result );
echo $row['webID'];
';
fwrite($filehandle, $datawrite);
fclose($filehandle);
You should escape the ' in the query with \' This will not exec and end the string
$filename = "websites/".$firstpage.".php";
$filehandle = fopen($filename, 'w') or die("error opening file");
$datawrite = '
$query = "SELECT websitetitle FROM $weblist WHERE webID=\'$webID\'";
$result = mysql_query ($query);
$row = mysql_fetch_array( $result );
echo $row[\'webID\'];
';
fwrite($filehandle, $datawrite);
fclose($filehandle);
Check this out for reference
http://php.net/manual/en/language.types.string.php
Just use proper escaping and place a \ before the apostrophe.
$datawrite = '
$query = "SELECT websitetitle FROM $weblist WHERE webID=\'$webID\'";
$result = mysql_query ($query);
$row = mysql_fetch_array( $result );
echo $row[\'webID\'];
';