json_encode() escaping / with \ [duplicate] - php

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json_encode() escaping forward slashes
(4 answers)
Closed 6 years ago.
I am trying to get html in a PHP array. When I print the variable out the table is in the correct format but be it is outputted to the array there are added backslashes to my html closing tags (Example: </th>). I didnt think there needed to be any escape characters for slash in PHP. When I print_r($table); it outputs without adding the backslash and outputs the table correctly in my browser but when I add the html table to a PHP array and then output the array as a JSON object it adds the backslash to my html closing tag as shown in the output below from my browser. Any thoughts would be helpful.
Output:
{"data":{"success":"true","carriers":"1,2","table":"
CarrierId<\/th>
CarrierName<\/th>
1<\/td> UHC<\/td><\/tr>
2<\/td> BlueCross<\/td><\/tr><\/table>"}}
PHP:
<?php
// Show all information, defaults to INFO_ALL
//phpinfo();
/*environment:
OS: windows Server 2012 Standard build 9200
IIS: IIS version 8.0.9200.16384 --> Authentication - win auth ->enabled, Anonymou auth -->disabled, CGI -> Impersonate User = true
http://www.microsoft.com/web/downloads/platform.aspx
http://blogs.msdn.com/b/brian_swan/archive/2010/02/10/sql-server-driver-for-php-understanding-windows-authentication.aspx
SQL: - MS SQL server 10.5.1600
PHP: PHP version 5.4.26 for IIS, copy php_sqlsrv_54_ts.dll to C:\Program Files (x86)\PHP\v5.4\ext directroy from
http://www.iis.net/learn/application-frameworks/install-and-configure-php-on-iis/install-the-sql-server-driver-for-php
php.ini file add
extension=php_pdo_sqlsrv_54_ts.dll
extension=php_sqlsrv_54_ts.dll
*/
//Connect to SQL with Authentication
$serverName = "PFIT-00-lync-03"; //serverName\instanceName
$connectionInfo = array( "Database"=>"spicewebinsurance", "UID"=>"spicerest", "PWD"=>"Password1");
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn ) {
//echo "Connection established.<br />";
}else{
echo "Connection could not be established.<br />";
die( print_r( sqlsrv_errors(), true));
}
$tsql= "select convert(varchar,CarrierId) as CarrierId, CarrierName as CarrierName from dbo.Carrier ";
//print_r($tsql) ;
$stmt=sqlsrv_query($conn, $tsql);
$table = "<table> <th>CarrierId</th><th>CarrierName</th>";
// Create table body
if ($stmt) {
$rows = sqlsrv_has_rows( $stmt );
if ($rows === true)
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) )
{
$carriers[] = $row['CarrierId'];
$table .= "<tr><td>";
$table .= $row['CarrierId'];
$table .= "</td><td>";
$table .= $row['CarrierName'];
$table .= "</td></tr>";
}
$table .= "</table>";
$carriers=implode(',', $carriers);
$return=array(
"success"=>"true",
"carriers"=>$carriers,
"table"=>$table
);
}
else
{
$return['success']=false;
}
echo '{"data":'.json_encode($return).'}';
sqlsrv_free_stmt( $stmt);
?>

Answer 1

I haven't read up on why the / is escaped but use option JSON_UNESCAPED_SLASHES:
echo '{"data":'.json_encode($return, JSON_UNESCAPED_SLASHES).'}';