I have several drop lists where if no option is selected then the value is = ""...
I cant figure out how to build the query for mysql in PHP.
query = SELECT * FROM db
I assume you have a select like this:
<select name="data[]" multiple="multiple">
<option value="A">A</option>
<option value="B">B</option>
<option value="C">C</option>
</select>
Your php can be something like
<?php
$data = array();
$data = $_POST['data'];
$query = "select * from table";
if (count($data > 0)) {
for ($i = 0; $i < count($data); $i++) {
$data[$i] = "'{$data[$i]}'";
}
$query .= " where field in (".implode(",", $data).")";
}
Too less information, but here's what I would do
$rows = $db->query(
'select *
from
table
where
checkbox_value = ?',
$_POST['checkbox']
);
In $rows you will have all the data you need.
You can run a SELECT on a table not on a DB! A Database consists of many tables. See http://www.php.net/manual/en/function.mysql-select-db.php
Check out w3Schools sql tutorials.
Or more specifically the select tutorial
Also the PHP/mysql tutorial will give you all that you need for this stuff.
Related
I want to write an IF statement for a dynamically generated dropdown menu, but I keep getting errors or it doesn't just work, what I am trying to do is compare two variables to see if they match from a data in the database
$loma = "Asokoro";
$row['locales'] is from locality table in the database
$row['locales'] = "Asokoro";
$row['locales'] = "Dutse";
$row['locales'] = "Mari";
$row['locales'] = "Cook";
That means if $loma which is Asokoro matches $row['locales'] = "Asokoro"; select it as the option menu
<select name="checkout_area_name" id="checkout_area_name" required>
$query = "SELECT * FROM `locality` WHERE state_name = '$hi_state'";
$sql = mysqli_query($con, $query) or die(mysqli_error($con, $query));
$r = ' <option value="">Please Choose Locality</option>';
?>
<?php
while ( $row = mysqli_fetch_assoc($sql))
{
?>
<?php $r = $r . '<option value="'.$loma.'" if ("'.$loma.' == '.$row["locales"].'") selected="selected" >'.$row['locales'].'</option>'; ?>
<?php
}
echo $r;
?>
</select>
I am trying to select the options menu that has $loma and $row['locales'] matching but I keep getting errors or when I console.log, it does not produce the result i want.
You are outputting php script as html markup, try changing your code to:
<select name="checkout_area_name" id="checkout_area_name" required>
$query = "SELECT * FROM `locality` WHERE state_name = '$hi_state'"; $sql = mysqli_query($con, $query) or die(mysqli_error($con, $query)); $r = '
<option value="">Please Choose Locality</option>'; ?>
<?php
while ( $row = mysqli_fetch_assoc($sql))
{
$r = $r . '<option value="'.$loma.'" '.(($loma==$row["locales"])?'selected':'').'>'.$row['locales'].'</option>';
}
echo $r;
?>
</select>
i need help for my php script with bootstrap selectpicker integration
i have php script for show selectpicker multiple value
<select id="skill" name="skill" class="selectpicker" multiple data-size="5" data-selected-text-format="count">
<?php
$skills = mysql_query("SELECT skill_id, skill_name FROM skill",$conn);
while($r_skills = mysql_fetch_assoc($skills)){
echo '<option value="'.$r_skills['skill_id'].'">'.$r_skills['skill_name'].'</option>
';
}
?>
</select>
and on my other table i have user and skill_id, for skill_id content : 1, 2, 3
i had explode it into each value like this :
$q = ("SELECT nik, skill_id FROM dosen WHERE nik = '".$nik."'",$conn);
while($row_dosen = mysql_fetch_assoc($q)){
$pieces = explode(",", $row_dosen['skill_id']);
for($i = 0; $i < count($pieces) ; $i++){
$pp = mysql_query("SELECT skill_id, skill_name FROM skill WHERE skill_id = '".$pieces[$i]."'",$conn);
while($p = mysql_fetch_assoc($pp)){
$var_each = $p['skill_id'];
}
}
}
The question is : how i can set default selected value for my multiple selectpicker option to make them selected as my user data on table ?
Thank you.
The help will be appreciated :)
How about something like:
<select id="skill" name="skill" class="selectpicker" multiple data-size="5" data-selected-text-format="count">
<?php
$skills = mysql_query("SELECT skill_id, skill_name FROM skill",$conn);
while($r_skills = mysql_fetch_assoc($skills)){
echo '<option value="'.$r_skills['skill_id'].'"';
if ($r_skills['skill_id'] == '1337 codingz')
echo ' selected';
echo '>'.$r_skills['skill_name'].'</option>';
}
?>
</select>
I am trying to search in MySql db based on user input something like below.
User may/not select from the below fields
<select name="field_one" id="f1">
<option value="AA">AA</option>
<option value="BB">BB</option>
<option value="CC">CC</option>
</select>
<select name="field_two" id="f2">
<option value="11">11</option>
<option value="22">22</option>
<option value="33">33</option>
</select>
if User selects only 'field_one', then mysql_query should filter only based on 'field_one'. In this case there can be four combination
Filed_one is slected and filed_two is not selected
Field_One is not selected and field_two is selected
Filed_one is not selected and Filed_two is not selected
Field_one is selected and Field_Two is selected
What is the best and efficient methode to make this search?
I tried with 'case .... break;' and 'if', but when the first condition is met, code stops and does not check the next conditions
Can any one give me a clue? Thanks in advance for the help....
try this:-
$query = "select * from table_name where 1 ";
if(!empty($_POST['field1']) ) {
$query .= " AND field1 like '".trim($_POST['field1'])."'";
}
if(!empty($_POST['field2'])) {
$query .= " AND field2 like '".trim($_POST['field2'])."'";
}
// and so on
$result = mysql_query($query);
please use escape string also http://php.net/mysql_real_escape_string
<?php
$sql = "SELECT * FROM table_name WHERE 1";
if(isset($_POST)){
if(isset($_POST['field_one'])){
$sql.= 'AND field_one'= $_POST['field_one'];
}
if(isset($_POST['field_two'])){
$sql.= 'AND field_two'= $_POST['field_two'];
}
}
mysql_query($sql);
?>
Example, not tested and needs lots of variable processing against SQL injection
$where = "";
$bits = array();
$bitset=false;
if(isset($_POST['field_one') && strlen($_POST['field_one')) > 0)
{
$bitset = true;
$bits[] = " field1 = $_POST['field_one')"
}
if(isset($_POST['field_two') && strlen($_POST['field_two')) > 0)
{
$bitset = true;
$bits[] = " field2 = $_POST['field_two')"
}
if($bitset)
{
$where = implode(", " $bits);
}
$sql = "select * from table " . $where;
You can also use PDO & param binding to avoid SQL Injection : http://www.php.net/manual/fr/pdostatement.bindparam.php
I have a form with a select field:
<select class="indexSearchLocationList" name="locationList">
<option value="allLocations">Anywhere in London</option>
<option value="barking_and_dagenham">Barking and Dagenham </option>
<option value="barnet">Barnet</option>
What im trying to do is if the user chooses the option allocations (Anywhere in London) then I need to use that to select all when querying the database below is the php I currently have but how do I specifiy if that particular option is choosen:
$choosenLocation = $_POST['locationList'];
Just try this, if it is not meet your requirement, then add your requirement briefly as comment.
$choosenLocation = $_POST['locationList'];
if($choosenLocation == 'allLocations')
{
$query = "select * from TABLE ";
}
else
{
$query = "select * from TABLE WHERE condition="any_condition";
}
I think that may be
if(isset($_POST['locationList']) & $_POST['locationList']=='allLocations')
{
$query = "select * from TABLE ";
//based on the mysqli_ or PDO you execute
//fetch the row and display accordingly
}
Do you expect something similar,
$choosenLocation = $_POST['locationList'];
if(isset($choosenLocation) && $choosenLocation == 'allLocations'){
$query = "select * from TABLE ";
//based on the mysqli_ or PDO you execute
//fetch the row and display accordingly
}else{
$query = "select * from TABLE WHERE condition='your condition here' " ;
}
I have a working mysql query that retrieves data from table1.
Now I will add every month a new table (table2, table3..etc).
Goal 1 : I would like to add a drop down menu and populate it with all tables as user options.
Goal 2 : I would like to make the query connect to the table that the user selects from the drop down menu, retrieves data and then refresh the page or just the table's div to display updated data.
my current mysql/php code :
$query = "SELECT X, Y, Z FROM **table1**";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
echo $row['X'];
echo $row['Y'];
echo $row['Z'];
}
Instead of "table1" it should be a variable linked to the drop down menu that the user selects.
I feel it should be simple, but I am bit new to this and could not find a similar case.
Thanks a lot gents.
I like the comment above but here is an example not sure if that what you are looking for
<form action="process.php" method='post'>
<select name="tables">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
<input type="submit" />
</form>
process.php file
$table=$_POST['tables'];
$query = "SELECT X, Y, Z FROM ".$table;
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
echo $row['X'];
echo $row['Y'];
echo $row['Z'];
}
$result = 'SHOW TABLES [{FROM | IN} db_name] [LIKE 'pattern' | WHERE expr];';
while($row = mysql_fetch_array($result))
{
echo $row['Tables_from_db_name'];
}