Mysqli_query inside of a foreach loop - php

I'm having trouble with a mysqli_query from inside a foreach loop, I'm getting a string from a table first, then separating that into an array. Then I try looping through the array and calling a query inside the loop.
$langs_result = mysqli_query($con, "SELECT Languages FROM users WHERE Username = '$username'");
$row = mysqli_fetch_assoc($langs_result);
$langs = $row['Languages'];
$userLangs = str_replace(" ","",$langs);
$userLangs = explode(",",$langs);
print_r($userLangs);
$posts = array();
foreach($userLangs as $lang){
echo "$lang <br>";
$sql = "SELECT * FROM posts WHERE Language = '$lang'";
$getLangPosts = mysqli_query($con, $sql);
array_push($posts, mysqli_fetch_assoc($getLangPosts));
}
print_r($posts);
for this user the langusges are German, Italian, Danish, and English, but the $posts array only contains the first post found from the first language (German), can anybody help? I am trying to get all of the posts for each language in the $userLangs array.
It's going through the foreach loop okay as the $lang variable that's echoed changes each time but the query still isn't working properly.
Thanks for the help!

select posts.* from posts
left join users on users.language=posts.language
where users.username='your_desiredusername'
group by users.language;
Just try to run this as a single query by filling the username
no need of multiple queries

You an avoid multiple queries by doing a JOIN, using FIND_IN_SET to match on your comma separated list. You probably need to use REPLACE to get rid of the extra spaces in the comma separated list as well.
Then you can just loop around to display the data, displaying the language on change of language:-
<?php
$sql = "SELECT a.Languages AS user_languages,
b.*
FROM users a
LEFT OUTER JOIN posts b
ON FIND_IN_SET(b.Language, REPLACE(a.Languages, ' ', ''))
WHERE a.Username = '$username'
ORDER BY b.Languages";
$langs_result = mysqli_query($con, $sql);
if($row = mysqli_fetch_assoc($langs_result))
{
print_r(explode(', ', $row['user_languages']));
$prev_langauge = '';
while($row = mysqli_fetch_assoc($langs_result))
{
if ($prev_langauge != $row['Languages'])
{
if ($prev_langauge != '')
{
print_r($posts);
}
$posts = array();
echo $row['Languages']."<br>";
$prev_langauge = $row['Languages'];
}
array_push($posts, mysqli_fetch_assoc($row));
}
if ($prev_langauge != '')
{
print_r($posts);
}
}

UPDATE
See this code:
<?php
$langs_result = mysqli_query($con, "SELECT Languages FROM users WHERE Username = '$username'");
$row = mysqli_fetch_assoc($langs_result);
$langs = $row['Languages'];
// $langs = 'German, Italian, Danish, French'; added this to run the test
// $userLangs = str_replace(" ","",$langs); this is not needed, see the explode below
$userLangs = explode(", ",$langs);
foreach($userLangs as $lang){
echo $lang;
$sql = "SELECT * FROM posts WHERE Language = '$lang'";
$getLangPosts = mysqli_query($con, $sql); // this is the result of the select *
while($post = mysqli_fetch_assoc($getLangPosts)){ // iterate over all the results
$postField = $post["yourChoiceField..say..Title"]; // get something from each row
array_push($posts, $title); // push into array
}
}
print_r($posts);
Since the initial select is based on username I don't believe the first loop is needed so your code was on the right track.
A second loop was needed to iterate over the posts though and a field to populate the $posts array.
You need to perform mysqli_fetch_assoc in a loop
$langs_result = mysqli_query($con, "SELECT Languages FROM users WHERE Username = '$username'");
while($row = mysqli_fetch_assoc($langs_result)){
$langs = $row['Languages'];
$userLangs = str_replace(" ","",$langs); // i don't get why you are doing this though
$userLangs = explode(",",$langs);
print_r($userLangs);
$posts = array();
foreach($userLangs as $lang){
echo "$lang <br>";
$sql = "SELECT * FROM posts WHERE Language = '$lang'";
$getLangPosts = mysqli_query($con, $sql);
array_push($posts, mysqli_fetch_assoc($getLangPosts));
}
print_r($posts);
}
It would help to know how what the select query actually returns.

mysqli_fetch_assoc only fetches one row on each call you need to use it like this:
while ($row_new = mysqli_fetch_assoc($getLangPosts)){
array_push($posts, $row_new);
}

You need to loop the inner query to get the column data
foreach($userLangs as $lang){
echo "$lang <br>";
$sql = "SELECT * FROM posts WHERE Language = '$lang'";
$getLangPosts = mysqli_query($con, $sql);
while($row1 = mysqli_fetch_assoc($getLangPosts))
array_push($posts, $row1['YOUR_COLUMN_NAME']);
}
OR you should use IN clause instead of loop
$langs_result = mysqli_query($con, "SELECT Languages FROM users WHERE Username = '$username'");
while($row = mysqli_fetch_assoc($langs_result)){
$langs = $row['Languages'];
$userLangs = str_replace(" ","",$langs);
$userLangs = explode(",",$langs);
print_r($userLangs);
$posts = array();
$sql = "SELECT * FROM posts WHERE Language IN ('".implode(',',$userLangs)."')";
$getLangPosts = mysqli_query($con, $sql);
while($row1 = mysqli_fetch_assoc($getLangPosts))
array_push($posts, $row1['YOUR_COLUMN_NAME']);
}

Related

Get All Rows When Variable Empty PHP Mysql

$frame_type = '';
$ret = mysqli_query($con, "select * from products where status='1' AND frame_type = '$frame_type' ");
while ($row = mysqli_fetch_array($ret)) {
$emparray[] = $row;
}
Get All Rows If The $frame_type Is Empty I am trying this way but i get zero rows , How to fix that Where $frame_type has value then send to query else not
There are a some of things that is wrong with your question (code). But if you only want answer. Just copy this
$frame_type = '';
$query = "select * from products where status='1' ";
// strlen has value
if(strlen($frame_type)) {
$query .= "AND frame_type = '$frame_type'";
}
$ret = mysqli_query($con, $query);
while ($row = mysqli_fetch_array($ret)) {
$emparray[] = $row;
}
PS: It's never a safe idea to pass everything to the query. Use prepared statement if you can.

Display single column value of mysqli query

How can I get a single column value from mysqli? The result should be single row with only one column.
This is what I have tried:
$query = "SELECT MAX(`userid`) FROM `user`";
$rlt = mysqli_query($this->db, $query);
echo $rlt['userid'];
You are not fetching the row after executing the query:
$query = "SELECT MAX(`userid`) FROM `user'";
$rlt = mysqli_query($this->db,$query);
$row = mysqli_fetch_row($this->db, $rlt);
echo $row[0];
The alternative would be to use an alias for the computed field and use fetch_assoc:
$query = "SELECT MAX(`userid`) as `maxid` FROM `user'";
$rlt = mysqli_query($this->db,$query);
$row = mysqli_fetch_assoc($this->db, $rlt);
echo $row['maxid'];
try with create alias and fetch result after query execution also your quote of user' looking wrong
$query = "SELECT MAX(`userid`) as userid FROM `user`";
$rlt = mysqli_query($this->db,$query);
$r = mysqli_fetch_assoc($rlt);
echo $r['userid'];
or fetch only one row like:-
$r = mysqli_fetch_row($this->db, $rlt);
echo $r[0];
You should create alias here.
Use this one:
$query = "SELECT MAX(`userid`) as Max_Userid FROM `user'";
$rlt = mysqli_query($this->db,$query);
$row = mysqli_fetch_assoc($this->db, $rlt);
echo $row['Max_Userid'];
Note: Always use alias when you use mysql function in query with fields.

How can i join or implode the below array code?

In the below code i want to join or implode all arrays of $trackersurl in a single line. i am getting the results in different lines, so i want to join in a single line.
Can anyone help me out?
I am searching results in stackoverflow, but could not follow.
My code is in below:
$sql = "SELECT * FROM announce WHERE torrent = $id ORDER BY seeders DESC";
$query = #mysql_query($sql);
while ($result = #mysql_fetch_array($query)) {
$trackersurl1 = $result['url'];
$trackersurl2 = "&tr=".$trackersurl1;
$trackersurl = array($trackersurl2);
}
Results of [var.trackersurl] in html page is below:
&tr=http:ajgdsjhg/ann
&tr=udp://iuysidfu/ann
&tr=udp:wutefghgw/ann
&tr=http://sdhgsjdhgj/ann
I want to join them in a single line below
&tr=http:ajgdsjhg/ann&tr=udp://iuysidfu/ann&tr=udp:wutefghgw/ann&tr=http://sdhgsjdhgj/ann
You should be careful of sql injection.
Are you looking to create an array['trackers'] with a string of all the trackers for a magnet link?
<?php
$sql = "SELECT * FROM announce WHERE torrent = ".mysql_real_escape_string($id)." ORDER BY seeders DESC";
$query = mysql_query($sql);
$tracker = null;
if(mysql_num_rows($query)>=1){
while ($result = mysql_fetch_array($query)) {
$tracker .= "&tr=".$result['url'];
}
}
$tracker = array('trackers'=>$tracker);
//$tracker['trackers'] = "&tr=a.com&tr=b.com&tr=c.com";
?>
Try this code
$newArray=array();
while ($result = #mysql_fetch_array($query)) {
$trackersurl1 = $result['title'];
$newArray[] = "&tr=".$trackersurl1;
}
$urlString=implode('',$newArray);

Creating an array of IDs from a while loop

Im trying to generate an array but not sure how to go about it.
I'm currently getting my data like so:
$query = mysql_query("SELECT * FROM users WHERE userEmail LIKE 'test#test.com'");
$row = mysql_fetch_array($query);
$query1 = mysql_query("SELECT * FROM categories");
while($row1 = mysql_fetch_array($query1)){
$query2 = mysql_query("SELECT * FROM usersettings WHERE userId = ".$row['userId']." AND usersettingCategory".$row1['categoryId']." LIKE 'y'");
$isyes = mysql_num_rows($query2);
if($isyes > 0){
$cat1 = mysql_query("SELECT * FROM shops WHERE shopstateId = 1 AND (categoryId1 = ".$row1['categoryId']." OR categoryId2 = ".$row1['categoryId']." OR categoryId3 = ".$row1['categoryId'].")");
$cat1match = mysql_num_rows($cat1);
if($cat1match > 0){
while($cat1shop = mysql_fetch_array($cat1)){
$cat1msg = mysql_query("SELECT * FROM messages WHERE shopId = ".$cat1shop['shopId']." and messagestateId = 1");
while($cat1msgrow = mysql_fetch_array($cat1msg)){
echo $cat1msgrow['messageContent']." - ".$cat1msgrow['messageCode'];
$cat1img = mysql_query("SELECT shopimagePath FROM shopimages WHERE shopimageId = ".$cat1shop['shopimageId']);
$imgpath = mysql_fetch_array($cat1img);
echo " - ".$imgpath['shopimagePath']."<br/>";
}
}
}
}
}
But this can cause duplicates when a user has all 3 of a shops categories picked in their preferences. I am trying to find a way to just pull the message ID out instead of the whole thing and put it into an array giving me, for example:
1,3,5,7,1,3,5,2,4,7,8
Then I can just run a separate query to say get me all messages where the ID is in the array, but i am unsure of the most constructive way to build such an array and examples of array from a while loop I have seen do not seem to be what I am looking for.
Is there anyone out there that can push me in the right direction?
Can't help with this code. But if you want an array from a query without duplicate result, you can use " select DISTINCT (id) " in your query or for more simple solution :
$id_arr = array();
$sql = mysql_query("select id from id_table");
while ($id_result = mysql_fetch_array($sql) {
$id = $id_result['id'];
if (!in_array($id, $id_arr)) {
$id_arr[] = $id;
}
}
I have found a much easier way to create the required result. I think at 6am after a hard night coding my brain was fried and I was making things a lot more complicated than I needed to. A simple solution to my issue is as follows:
$query = mysql_query("SELECT * FROM users WHERE userEmail LIKE 'test2#test2.com'");
$row = mysql_fetch_array($query);
$categories = "(";
$query1 = mysql_query("SELECT * FROM categories");
while($row1 = mysql_fetch_array($query1)){
$query2 = mysql_query("SELECT usersettingCategory".$row1['categoryId']." FROM usersettings WHERE userId = ".$row['userId']);
$row2 = mysql_fetch_array($query2);
if($row2['usersettingCategory'.$row1['categoryId']] == y){
$categories .= $row1['categoryId'].",";
}
}
$categories = substr_replace($categories ,")",-1);
echo $categories."<br />";
$query3 = mysql_query("SELECT * FROM shops,messages WHERE shops.shopId = messages.shopId AND messages.messagestateId = 1 AND (shops.categoryId1 IN $categories OR shops.categoryId2 IN $categories OR shops.categoryId3 IN $categories)");
while($row3 = mysql_fetch_array($query3)){
$query4 = mysql_query("SELECT shopimagePath FROM shopimages WHERE shopimageId = ".$row3['shopimageId']);
$row4 = mysql_fetch_array($query4);
echo $row3['messageContent']." - ".$row3['messageCode']." - ".$row4['shopimagePath']."<br />";
}

problem of while loop and array

$result=array();
$table_first = 'recipe';
$query = "SELECT * FROM $table_first";
$resouter = mysql_query($query, $conn);
while ($recipe = mysql_fetch_assoc($resouter, MYSQL_ASSOC)){
$result['recipe']=$recipe;
$query2="SELECT ingredients.ingredient_id,ingredients.ingredient_name,ingredients.ammount FROM ingredients where rec_id = ".$recipe['rec_id'];
$result2 = mysql_query($query2, $conn);
while($ingredient = mysql_fetch_assoc($result2)){
$result['ingredient'] = $ingredient;
}
echo json_encode($result);
}
this code show me all the recipes but only the last ingredients i.e
{"recipe":{"rec_id":"14","name":"Spaghetti with Crab and Arugula","overview":"http:\/\/www","category":"","time":"2010-11-11 14:35:11","image":"localhost\/pics\/SpaghettiWithCrabAndArugula.jpg"},"ingredient":{"ingredient_id":"55","ingredient_name":"test","ammount":"2 kg"}}{"recipe":{"rec_id":"15","name":"stew recipe ","overview":"http:\/\/www","category":"","time":"2010-11-11 14:42:09","image":"localhost\/pics\/stew2.jpg"},"ingredient":{"ingredient_id":"25","ingredient_name":"3 parsnips cut into cubes","ammount":"11"}}
i want to output all the ingredient records relevant to recipe id 14 and this just print the last ingredient.
$result['ingredient'] = $ingredient;
Is replacing the variable $result['ingredient'] with the most recent $ingredient value each time, culminating with the last value returned, you should use:
$result['ingredient'][] = $ingredient;
To incrememnt/create a new value within the $result['ingredient'] array for each $ingredient. You can then output this array according to your needs. Using print_r($result['ingredient']) will show you its content...to see for yourself try:
while($ingredient = mysql_fetch_assoc($result2)){
$result['ingredient'][] = $ingredient;
}
print_r($result['ingredient']);
If I understand correctly what you want to do, there is a way to optimize the code a lot, by fetching recipes and ingredients in one single query using a JOIN :
$recipes = array();
$recipe_ingredients = array();
$query = "SELECT * FROM recipe LEFT JOIN ingredients ON ingredients.rec_id=recipe.rec_id ORDER BY recipe.rec_id DESC";
$resouter = mysql_query($query, $conn);
$buffer_rec_id = 0;
$buffer_rec_name = "";
$buffer_rec_cat = "";
while( $recipe = mysql_fetch_array($resouter) )
{
if( $buffer_rec_id != $result['recipe.rec_id'] )
{
$recipes[] = array( $buffer_rec_id, $buffer_rec_name, $buffer_rec_cat, $recipe_ingredients);
$recipe_ingredients = array( );
$buffer_rec_id = $result['recipe.rec_id'];
$buffer_rec_name = $result['recipe.rec_name'];
$buffer_rec_cat = $result['recipe.rec_category'];
}
else
{
$recipe_ingredients[] = array( $result['ingredient_id'], $result['ingredient_name'], $result['ammount'] );
}
}
$print_r($recipes);
This code should give you a multi-dimensional array that you can use later to get the data you want.
I let you adapt the code to have exactly the information you need.

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