PHP Retrieve results - php

I am having a small trouble retrieving results that I hope someone can help me with.
I have a field called $incategory which is a comma based string, and what I want to do is explode the into an array that can be used to retrieve results as below (Hope that makes sense):
<?php
$showlist = $row_listelements['incategory'];
// ** e.g. $incategory = 1,3,5,
// ** What I want to do is look at table 'category'
// ** and retrieve results with an 'id' of either 1, 3 or 5
// ** Display Results
mysql_select_db($database_db, $db);
$query_display = "SELECT * FROM category WHERE id = ".$showlist." ORDER BY name ASC";
$display = mysql_query($query_display, $db) or die(mysql_error());
$row_display = mysql_fetch_assoc($display);
$totalRows_display = mysql_num_rows($display);
?>

You can use the IN keyword of SQL directly like this.
query_display = "SELECT * FROM category WHERE id IN (".$showlist.") ORDER BY name ASC";
Another tip would be to stop using MYSQL_QUERY as it is deprecated in PHP 5.3
Edit: If $showlist = '1,3,5,' you will need to remove the last comma from the string to make it useable in the query. Just use this query then
query_display = "SELECT * FROM category WHERE id IN ('".str_replace(",", "','", substr($showlist, -1))."') ORDER BY name ASC";

Use explode function and use , as delimiter.
refer here http://www.w3schools.com/php/func_string_explode.asp
Hope this helps.

First, you have explode the $incategory string into an array containing all of the category number. For example:
$incategory = explode(",", $incategory);
And then you just have to execute this query:
$query_display = "SELECT * FROM category WHERE id = "
. $incategory[$i] . " ORDER BY name ASC";
The $i should be defined beforehand (usually using loop).

Related

Select data from the database before or after a defined value

I am having problems achieving the query to select data from a table in the db after a defined value has been met.
My code to do this is:
$fi = 'first_segment'
$im = popo.jpg
$sqls = "SELECT * FROM $fi,news_pictures
WHERE $fi.pi_id = news_pictures.pi_id
AND news_pictures.i_name = '$im'
GROUP BY news_pictures.id DESC";
I wasn't able to achieve the result with that query.
Basically, I want the query to confirm if news_pictures.i_name = '$im' and if true, return starts from the value of $im followed by other data in the table depending on news_pictures.id DESC.
The sample data and output:
Table news_pictures:
id i_name
-- ------
1 coco.jpg
2 lolo.jpg
3 popo.jpg
4 dodo.jpg
Since $im = popo.jpg, I want my query to display all values starting from popo.jpg with id DESC, i.e. popo.jpg, lolo.jpg, coco.jpg.
I got to solve the question with the help of a friend.
$fsqls = "SELECT * FROM $fi,news_pictures WHERE $fi.pi_id = news_pictures.pi_id AND news_pictures.i_name = '$im' GROUP BY news_pictures.id";
$rres = mysqli_query($connection, $fsqls) or print(mysqli_error($connection));
while($row = mysqli_fetch_assoc($rres))
{
$rnm = $row["id"];
}
$sqls = "SELECT * FROM news_pictures WHERE news_pictures.id <= $rnm ORDER BY news_pictures.id DESC";

How to create multiple word search? SQL

We have made a search field where you can search for ingredients and find recipes.
For now you can only type in 1 ingredient:
if (isset($_POST['search'])) {
$searchquery = $_POST['search'];
$query = mysql_query("SELECT * FROM opskrifter WHERE id IN
(SELECT opskrifterid FROM ingredienser WHERE ing_name IN ('$searchquery'))") or die("search failed");
We want to be able to search for multiple ingredients in the same search field by seperating the ingredients with a "," or something like this.
Is there a simple way to make that happen ?
EDIT:
We tried to use explode like this without succes.
$searchTerms = explode(' ', $searchquery);
$searchTermBits = array();
foreach ($searchTerms as $term) {
if (!empty($term)) {
$searchTermBits[] = "ing_name '$term'";
}}
...
$result = mysql_query("SELECT * FROM opskrifter WHERE id IN (SELECT * FROM WHERE ".implode(' AND ', $searchTermBits)));
Thanks! :)
You could simply get the user to type in his values comma-separated, the the input would be almost in the right syntax for the query. You just have to add semicolons around the values because you search for a string in your table.
You can use PHP's str_replace()-Function:
$vals = $_POST['search'];
$valsFormatted = "'" . str_replace(",", "','", $vals) . "'";
In this code, you replace all the commas of the input with the comma plus semicolons before and behind them in orderto wrap all values of the input with semicolons. You also have to add one at the beginning and at the end of the string. Replace the first comma in the function above with the char you want your users to separate the values with.
After that, you can simply change your query to the following:
$query = "SELECT * FROM opskrifter WHERE id IN
(SELECT opskrifterid FROM ingredienser WHERE ing_name IN ('$valsFormatted'))";
Please also be informed, that your code like this is vulnerable for SQL Injections! Check out this link to learn how to prevent this.
A simple statement like this would work:
$array = implode("','",explode($_POST['search'], ","));
$query = mysql_query("SELECT * FROM opskrifter WHERE id IN (SELECT opskrifterid FROM ingredienser WHERE ing_name IN ({$array}))") or die("search failed");
First explode your search, then implode it (might not even need to do so). After that make sure the array gets used as the 'in' operator as a string/array.
For more information about this, you could read this question: PHP/MySQL using an array in WHERE clause
The working copy from my local machine was this;
$_POST['search'] = "0, 1, 2";
$array = implode ( "','", explode ( ",", $_POST['search'] ) );
$query = mysql_query("SELECT * FROM users WHERE id IN ('$array')") or die(mysql_error());
var_dump ( $array );
var_dump ( $query );
var_dump ( "SELECT * FROM users WHERE id IN ('$array')" );
var_dump ( mysql_fetch_array ( $query ) );
which actually did return users, so if we would take this example and change it to your code, it would be (the query, at least):
$query = mysql_query("SELECT * FROM opskrifter WHERE id IN (SELECT opskrifterid FROM ingredienser WHERE ing_name IN ('$array'))") or die(mysql_error());
Do take note of the changed $array variable too.
First you need to convert the text coming from the search field to array with:
$string = $_POST['search'];
$array = explode( '"' , $string);
So if you put in the search: test"hello"hi
the array will be:
1 => test,
2 => hello,
3 => hi
After that, you need to use the SQL format:
WHERE column_name IN (value1,value2,...)
So you need to change the array we have created to a string with this format:
$string = implode(',',$array);
So the echo of $string will be:
test,hello,hi
and SQL will be :
WHERE column_name IN ($string)

How to return only items that occur in 2 sql select statemnts

I have two different sql statements. $sql grabs all the items whose title matches a certain search text. $cat_sql grabs all the category_items that are in a certain category. An item has an ID. A category_item has a field called item_id which is a foreign key to IDs in the items table
...
mysqli setup code
...
$title = $_POST["title"];
$cat_id = $_POST["cat_id"];
$cat_sql = "SELECT * FROM category_items WHERE category_id = '".$cat_id."'";
$sql = "SELECT * FROM items where title LIKE '%". $title ."%' Limit 70";
if (!$result_cat = $mysqli->query($cat_sql)) {
// The query failed.
echo "<h2 >ERROR</h2>";
exit;
}
if (!$result = $mysqli->query($sql)) {
// The query failed.
echo "<h2 >ERROR</h2>";
exit;
}
Then I display all items:
while ($item = $result->fetch_assoc()) {
include 'item_card.php';
}
Currently this just displays all items fetched in the $sql query. Is there some way to remove all items from $result that do not have their ID represented as an item_id in $result_cat?
NOTE:
I would strongly prefer not to do just combine both SELECT statements into a table join because the actual $sql and $cat_sql are not nearly as simple as I have represented here. Also, they vary depending on which if statement they are in.
My question is: given $result and $result_cat, can I remove items from $result?
EDIT 1
As suggested by comments I am making an array if item_ids then doing an in_array query. Progress thus far:
$result_cat_ids = [];
while ($cat_item = $result_cat->fetch_assoc()) {
$result_cat_ids[] = $cat_item['item_id'];
}
EDIT 2 Here is the working code following the suggestions in the comments
if (in_array($item['id'], $result_cat_ids)) {
include 'item_card.php';
}
You may also use 'INTERSECT' sql clause.
$sql = "SELECT * FROM items WHERE id IN (SELECT item_id FROM category_items WHERE category_id = '".$cat_id."' INTERSECT SELECT id FROM items where title LIKE '%". $title ."%')";
This way, you can query for items that accomplish both conditions.
Note: I'm not using "limit 70" but you may add it as well.

PHP: How to insert array value into mysql statement

i have this array and get it from from an url. this array is member id that i need to pass to mysql.
$member_id = $_GET['member_id'];
the array like this : Array ( [0] => 1269 [1] => 385 )
how can i transfer this array into my mysql statement and make , become AND :
$answer_sql = mysql_query("SELECT tna_category. * , tna_question. *, tna_answer. *
FROM tna_category, tna_question, tna_answer
WHERE tna_category.section_id = '$section_id1'
AND tna_question.id = tna_answer.question_id AND tna_question.category_id = tna_category.id
AND tna_answer.member_id = ['1269' , '385']
ORDER BY tna_answer.question_id");
should i put bracket?..
in this part : tna_answer.member_id = Array or $member_id
As others have said, you can use IN() but you are apparently open to SQL injection attacks as it is. You need to do this:
$escaped_ids = array_map('mysql_real_escape_string', $member_ids);
Or, if they are surely all integers
$escaped_ids = array_map('intval', $member_ids);
Then, you can write your query like:
$query = "SELECT tna_category. * , tna_question. *, tna_answer. *
FROM tna_category, tna_question, tna_answer
WHERE tna_category.section_id = '" . mysql_real_escape_string($section_id1) . "'
AND tna_question.id = tna_answer.question_id
AND tna_question.category_id = tna_category.id
AND tna_answer.member_id IN (".implode(",", $escaped_ids).")
ORDER BY tna_answer.question_id";
Never, never, never put unescaped values in your query.
Also, you should not be using the mysql_ functions anymore. Please consider using the mysqli_ functions instead.
First split the array value, get no. of rows in the array value and pass the value one by one into the query by using for or foreach loop.
try this
$member_id = $_GET['member_id'];
If you're already getting comma seprated values then there's no need to use explode function just use implode function in database query.
$member_id = explode(",", $member_id);
and then
answer_sql = mysql_query("SELECT tna_category. * , tna_question. *, tna_answer. *
FROM tna_category, tna_question, tna_answer
WHERE tna_category.section_id = '$section_id1'
AND tna_question.id = tna_answer.question_id AND tna_question.category_id = tna_category.id
AND tna_answer.member_id IN (".implode(",", $member_id).")
ORDER BY tna_answer.question_id");
the explode function create array it depends on you explode value with comma OR space and then implode mean join these values with comma OR space.
for more detail explode and implode.
you can use IN clause of mysql like this
$your_array = array("0"=>"1269", "1"=>"385");
$in_text = implode(",", $your_array);
$sql = "SELECT tna_category. * , tna_question. *, tna_answer. *
FROM tna_category, tna_question, tna_answer
WHERE tna_category.section_id = '$section_id1'
AND tna_question.id = tna_answer.question_id
AND tna_question.category_id = tna_category.id
AND tna_answer.member_id IN ($in_text)
ORDER BY tna_answer.question_id";

WHERE id=(array)

The code below only outputs single line(there are 2 in database that should be outputed).
I think that problem is in id=$data[id] since data1 is array instead of single value.I hoped that while will fix that but it doesnt look too good...
$results1 = mysql_query("SELECT * FROM keywords WHERE keyword='$search' ORDER BY (relevant-irrelevant) DESC");
$data1=mysql_fetch_array($results1);
$results2=mysql_query("SELECT * FROM searchengine WHERE id='$data1[id]'");
while($data2=mysql_fetch_array($results2))
First, isolate your ids, looping to get all of the results:
$ids = array();
while ( $data1 = mysql_fetch_array($results1) ) {
$ids[] = $data1['id'];
}
Then, convert your $ids array into a string. An easy way to do this is via implode():
$results2=mysql_query(
"SELECT * FROM searchengine WHERE id IN (" . implode(',', $ids) . ")"
);
Maybe I´m missing something, but how can $data1['id'] be an array? it´s probably an integer and perhaps a string, but it's not an array. $data1['id'] is a single value; the value of field id in the keywords table
I think you just need to put curly quotes around the variable:
$results2=mysql_query("SELECT * FROM searchengine WHERE id='{$data1[id]}'");
or even better:
$results2=mysql_query("SELECT * FROM searchengine WHERE id=" . (int) $data1['id']);
If id is an integer that is.
And of course if the first query returns more than 1 result, you will have to loop through them as well.
Couldn't you just select the entire thing in one query?
SELECT *
FROM keywords k
searchengine s
WHERE k.keyword='$search'
AND k.id = s.id
$results1 = mysql_query("SELECT * FROM keywords WHERE keyword='$search' ORDER BY (relevant-irrelevant) DESC");
$data1=mysql_fetch_array($results1);
//VERY DANGEROUS TO USE USER INPUT
$in = join(',',$data1['id']);
$results2=mysql_query("SELECT * FROM searchengine WHERE id IN ({$in})");
while($data2=mysql_fetch_array($results2))
You can't pass array as condition. You should:
a. do a for(each) loop in the $data1 array and perform next actions
b. implode the array and search with IN. Example:
$commaSeparated = implode(",", $data1);
$results2=mysql_query('SELECT * FROM searchengine WHERE id IN ('.$commaSeparated.'));
mads.ohm is correct about combining the two queries into a single query.
As for your problem with only getting one return value, your while loop is just overwriting the contents of $data2 each time through.
You could write something like this instead:
$i = 0;
$data2 = array();
while ($row = mysql_fetch_array($results2)) {
$data2[$i] = $row;
$i++;
}
In this case, $data2 is declared as an array, and each iteration of the while loop adds a row from the database to the array.

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