I'm using the following code in PHP.
Its creating the file post.json, but inserting values into the table is not reflecting.
<?php
$input = file_get_contents('php://input');
logToFile("post.json",$input);
$json_a = json_decode($input,true);
$con = mysql_connect("localhost","root","root");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("mydb", $con);
mysql_query("INSERT INTO order_table (Table_id, Menu_Item_Name, Menu_Item_Price, Menu_Item_quantity) VALUES('$json_a[TableId]','$json_a[ItemName]','$json_a[ItemPrice]','$json_a[ItemQuantity]')");
function logToFile($filename,$msg)
{
$fd = fopen($filename,"a");
$str="[".date("Y/m/d h:i:s")."]".$msg;
fwrite($fd,$str."\n");
fclose($fd);
}
?>
Try to replace '$json_a[foo]' with '" . $json_a['foo'] . "' or '${json_a['foo']}'.
Related
I do not understand what is going wrong with code. The result is get is "connected successfully success Query failed". I tried few combinations and I get the same result. Please help me in solving this. Thanks in advance.
<?php
$link = mysql_connect('localhost', 'root1', '')
or die('Could not connect: ' . mysql_error());
if ($link) {
echo 'connected successfully';
}
$l = mysql_select_db('vtflix', $link) or die ('Could not select the database');
if ($l) {
echo ' success';
}
/*$varCNAME = 'John';
$varCONTENT = '4';
$varVID = '1';*/
$sql = "INSERT INTO mpaa(C_Name, ContentRating, V_ID) VALUES ('Jon', 4, 3)";
mysql_query($sql, $link) or die("Query failed");
$que = "SELECT * FROM mpaa";
$query = mysql_query($que, $link);
if (!$query) {
echo 'query failed';
}
while ($sqlrow = mysql_fetch_array($query, MYSQL_ASSOC)) {
$row = $sqlrow['C_Name'];
$nrow = $sqlrow['Content Rating'];
$mrow = $sqlrow['V_ID'];
echo "<br>" . $row . " " . $nrow . " " . $mrow . "<br>";
}
mysql_close($link);
?>
1.Don't use mysql_* library (deprecated from php5 onward + removed from php7) .Use mysqli_* OR PDO.
2.An example of mysqli_*(with your code)is given below:-
<?php
error_reporting(E_ALL); // check all type of error
ini_set('display_errors',1); // display those errors
$link = mysqli_connect('localhost', 'root1', '','vtflix');
if($link){
echo 'connected successfully';
$sql= "INSERT INTO mpaa(C_Name,ContentRating,V_ID) VALUES ('Jon', 4, 3)";
if(mysqli_query($link,$sql)){
$query = "SELECT * FROM mpaa";
$res = mysqli_query($link,$query);
if($res){
while($sqlrow=mysqli_fetch_assoc($query))
{
$row= $sqlrow['C_Name'];
$nrow= $sqlrow['Content Rating'];
$mrow= $sqlrow['V_ID'];
echo "<br>".$row." ".$nrow." ".$mrow."<br>";
}
mysqli_close($link);
}else{
echo die('Query error: ' . mysqli_error($link));
}
}else{
echo die('Query error: ' . mysqli_error($link));
}
}else{
echo die('Could not connect: ' . mysqli_connect_error());
}
?>
Note:- To check php version (either on localhost or on live server) create a file with name phpInfo.php, and just write one line code in that file:-
<?php
phpinfo();
?>
Now run this file and you will get the current php version.
Like this:- https://eval.in/684551
Here it seems that you are using deprecated API of mysql_* .
1) Check your PHP version
<?php phpinfo();exit;//check version ?>
2) avoid the usage of mysql use mysqli or PDO
3) change your db connection string with this :
new Mysqlidb($hostname, $username, $pwd, $dbname);
example with you code
<?php
$link = mysqli_connect('localhost', 'root1', '','vtflix');
if($link){
echo 'connected successfully';
$sql= "INSERT INTO mpaa(C_Name,ContentRating,V_ID) VALUES ('Jon', 4, 3)";
if(mysqli_query($link,$sql)){
$query = "SELECT * FROM mpaa";
$res = mysqli_query($link,$query);
if($res){
while($sqlrow=mysqli_fetch_assoc($query))
{
$row= $sqlrow['C_Name'];
$nrow= $sqlrow['Content Rating'];
$mrow= $sqlrow['V_ID'];
echo "<br>".$row." ".$nrow." ".$mrow."<br>";
}
mysqli_close($link);
}else{
echo die('Query error: ' . mysqli_error($link));
}
}else{
echo die('Query error: ' . mysqli_error($link));
}
}else{
echo die('Could not connect: ' . mysqli_connect_error());
}
?>
When trying to post data to the database it gives an error saying no database selected. But the content from the database is displaying on a table and select menu.
<?php
$con = mysqli_connect("localhost", "root", "","radian");
if(!$con)
{
exit("Couldn't connect: ".mysqli_connect_error());
}
mysqli_set_charset($con, "utf8");
$insert_data = "UPDATE enquiries
SET ResponseDate = '".$current_date."',
Response = '".$txtResponse."',
Enquiry_No = '".$_SESSION['ses_staff']
."' WHERE Enquiry_No = '".$txtStudentId."'";
$execute = mysql_query($insert_data) or die(mysql_error());
$output= '<h4 style="margin-left:1em;width:15em;color:red;"> Response successful!. </h4>';
}else{
$output= '<h4 style="margin-left:1em;width:15em;color:red;"> </h4>';
}
Your final code should be identical to:
<?php
$con = mysqli_connect("localhost", "root", "", "radian");
if (!$con) {
exit("Couldn't connect: " . mysqli_connect_error());
}
mysqli_set_charset($con, "utf8");
$insert_data = "UPDATE enquiries SET ResponseDate = '" . $current_date . "', Response = '" . $txtResponse . "',Enquiry_No = '" . $_SESSION['ses_staff'] . "' WHERE Enquiry_No = '" . $txtStudentId . "'";
$execute = mysqli_query($con, $insert_data) or die(mysqli_error($con));
$output = '<h4 style="margin-left:1em;
width:15em;
color:red;"> Response successful!. </h4>';
Changes made:
Remove all instances of mysql_* and replace with the correct mysqli_* function.
Remove the orphaned } else {.
Note: Officially mysql_* functions are deprecated. So no point using them. Use either mysqli_* or PDO.
I'm newbie in PHP sorry. How can i update a user defined var in user defined table with user defined value in MySql? I cant get it work.
<?PHP
$table = $_POST['table'];
$id = $_POST['id'];
$key = $_POST['key'];
$value = $_POST['value'];
$con = mysql_connect("mysql.serversfree.com","u105645000***","mf***") or ("Cannot connect!" . mysql_error());
if (!$con)
die('Could not connect: ' . mysql_error());
mysql_select_db("u10564500***" , $con) or die ("could not load the database" . mysql_error());
/*$mysql = mysql_query("INSERT INTO $table ($key) VALUES ('".$value."');");*/
$sql = "UPDATE $table Set ".$key."='".$value."'";
?>
Why your code is not working
You're not launching the query.
No where clause
Without specifying any conditions, every record of the coulmn $key will be updated with $value.
Solution
Use this code:
<?PHP
$con = mysqli_connect("mysql.serversfree.com","u105645000***","mf***","u10564500***");
$table = mysqli_real_escape_string($con,$_POST['table']);
$id = mysqli_real_escape_string($con,$_POST['id']);
$key = mysqli_real_escape_string($con,$_POST['key']);
$value = mysqli_real_escape_string($con,$_POST['value']);
if (!$con){die('Could not connect: ' . mysqli_error($con));}
$sql = mysqli_query($con,"UPDATE ".$table." Set ".$key."='".$value."'");
?>
I am trying to import xml data into a mysql table. I have the below fields to be Imported:
reference
price
category
type
city
property
imgurl1
imgurl2
The problem is that the number of <imgurl>(url to image file) is not the same. Below is the code:
$conn = mysql_connect($hostname, $username, $password);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($dbname,$conn) or die( mysql_error() );
$xml = simplexml_load_file('http://astonpearlemail.net/feeds/feedsmall.xml');
$data = $fields = array();
foreach ($xml->xpath('listing') as $listing) {
$fields = array_keys((array)($listing));
$data[] = '("' . join('", "', (array)$listing) . '")';
}
$sql = "INSERT INTO ap_prop (" . join(', ', $fields) . ") VALUES\n" ;
$sql .= join (",\n", $data);
$result1 = mysql_query($sql,$conn);
echo "<pre>$sql</pre>"
Please suggest how I can import the varying number of imgurl.
Thanks in advance
I have the following code in php:
<?php
$con = mysql_connect("localhost","john","john");
mysql_select_db("data", $con);
if (!$mysql_select_db) {
die ('Cannot use data : ' . mysql_error());
}
$result = mysql_query("SELECT * FROM test where huss='hussein'");
while ($row = mysql_fetch_array($result)) {
echo "ADSDAD";
echo "<br />";
}
if (!$con) {
die('ERROR Could not connect: ' . mysql_error());
}
mysql_close($con);
?>
I was able to connect to the MySQL server, but I can't seem to select the relevant database.
Being printed is the hard-coded text Cannot use data :, but no MySQL error message follows it. If the MySQL connection had failed then I would have expected a MySQL error message to appear after the hard-coded text.
What am I doing wrong?
The problem's here:
mysql_select_db("data", $con);
if (!$mysql_select_db) {
die ('Cannot use data : ' . mysql_error());
}
You don't have a variable $mysql_select_db, and it doesn't magically map to the result of the last time you called mysql_select_db.
So, instead:
$result = mysql_select_db("data", $con);
if (!$result) {
die ('Cannot use data : ' . mysql_error());
}
Or:
if (!mysql_select_db("data", $con)) {
die ('Cannot use data : ' . mysql_error());
}