I have made database that contains some information. Want to do is to retrieve and display data that fits with parameters like:
if i select green from dropdown list, it will display only information about cars that are green. Also "kõik" means like all.
Here is form that im using.
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body align="center" style="margin-top:200px;">
<table border="1" align="center">
<tr>
<td>
<form action="retrieve.php" method="get">
Rass:
<select name="rass">
<option value="ebony">Ebony</option>
<option value="Valge">Valge</option>
<option value="Aasia">Aasia</option>
</select> <br />
<input type="submit" value="lisa" />
</form>
</td>
</tr>
</table>
</body>
</html>
Here is php that used to show all data.
<?php
$connect = #mysql_connect ("localhost", "root", "") or die("Fail!!!! :D:D:D");
mysql_select_db("tibid") or die("selline andmebaas puudub");
$query = mysql_query("select * from test");
$num_rows = mysql_num_rows($query);
if($num_rows > 0){
{while($row = mysql_fetch_assoc($query))
echo
$row['rinnad']. "<br>"
.$row['juuksed']."<br>"
.$row['silmad']."<br>"
.$row['rass']."<br>
<hr>
";}
}else{
echo "Andmebaas on tühi";
}
?>
Link to this kind tutorial would be helpful.
Thank you
I think you are looking for WHERE. So your query would be:
select * from cars where color = '$color'
where $color is escaped $_POST['color'] passed from your form.
Related
unable to list the information in LOV list from mysql table . I need to list the categories(catname) in the dropdown box and also let me know how to select the value(catno) for further action.
<?php
require_once('dbconnect.php');
$selsql="SELECT catno,catname FROM category"; //catno-integer and catname-
varchar
$res=mysqli_query($con,$selsql);
//$r=mysqli_fetch_assoc($res);
if (mysqli_query($con,$selsql)) {
}
else
{
echo "No connection";
die(mysqli_error($con));
}
?>
<html>
<head>
<title>Drop down list </title>
<meta charset=UTF-8">
<meta name="viewport">
</head>
<body>
<select name="categories"
style="width:250px;"onchange="this.form.submit();">
<?php while($row=mysqli_fetch_assoc($res)):;?>
<option value="<?php echo $row[0];?>"<?php echo $row[1];?></option>
<?php endwhile;?>
<!---<option value="<?php echo $row[0];?>""<?php echo $row[1];?>"</option>->
<!----$options.='<option value="'.$row[0].'"selected>'.$row[1].'</option>'->
</select>
</body>
</html>
This should work.
<?php
require_once('dbconnect.php');
$selsql="SELECT catno,catname FROM category";
$res=mysqli_query($con,$selsql);
if ($res == '')
{
echo "No connection";
die(mysqli_error($con));
}
?>
<html>
<head>
<title>Drop down list </title>
<meta charset="UTF-8">
<meta name="viewport">
</head>
<body>
<select name="categories" style="width:250px;" onchange="this.form.submit();">
<?php while($row=mysqli_fetch_assoc($res)) {?>
<option value="<?php echo $row['catno'];?>"><?php echo $row['catname'];?></option>
<?php } ?>
</select>
</body>
</html>
Hi guys i have made a listbox with data from a mysql database and now i want to give the user the possibility to insert an option that don't exists. Can anyone tell me how to do that? I want to create a form or another thing that allows the user to introduce a value for a new option and then it appears in listbox and forward get the value to save in mysql database.
Best regards.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<meta name="keywords" content="jquery,ui,easy,easyui,web">
<meta name="description" content="easyui help you build your web page easily!">
<link rel="stylesheet" type="text/css" href="jeasyui_src/themes/default/easyui.css">
<link rel="stylesheet" type="text/css" href="jeasyui_src/themes/icon.css">
<link rel="stylesheet" type="text/css" href="jeasyui_src/demo/demo.css">
<script type="text/javascript" src="jeasyui_src/jquery.min.js"></script>
<script type="text/javascript" src="jeasyui_src/jquery.easyui.min.js"></script>
</head>
<body>
<script language='Javascript' type='text/javascript'>
function edit_file()
{
$("#button_file").css("visibility" , "hidden");
$("#file_new").css("visibility" , "visible");
}
</script>
<h3>Coloque aqui a sua revisao tecnica:</h3></br>
<?php
include_once 'acess_db.php';
$query = "select * from faqs_treeview where level=1 order by category_title";
$result = mysql_query($query);
?>
<table border='0'>
<tr>
<td>
<form method="POST" name="form1" id="t1" style="visibility: visible;">
<select name="cat_1" style="visibility: visible;">
<option>Selecione a categoria</option>
<?php
while($row = mysql_fetch_array($result))
{
$id = $row["id_category"];
$name = $row["category_title"];
echo "<option value='$id'>".$name."</option>";
}
?>
</select>
<input type="submit" name="submit1" onclick="open2();">
</form>
<?php
if(isset($_POST["cat_1"]))
{
// echo $_POST["cat_1"];
$id2 = $_POST["cat_1"];
$query1 = "select * from faqs_treeview where high_level=$id2";
$result1 = mysql_query($query1);
$form_visible = "visible";
}
else
{
$form_visible = "hidden";
}
?>
</td>
<td>
<form method="POST" name="myform2" id="t2" style="visibility: <?= $form_visible ?>">
<select name="cat_2" >
<option>Selecione a sub-categoria</option>
<?php
while($row1 = mysql_fetch_array($result1))
{
$id = $row1["id_category"];
$name = $row1["category_title"];
echo "<option value='$id'>".$name."</option>";
}
?>
</select>
<input type="submit" name="submit2" onclick="open3();">
<input type="hidden" name="cat_1" value="<?= $_POST["cat_1"]?>">
</form>
</td>
<?php
//echo $_POST["cat_2"];
if(isset($_POST["cat_2"]) )
{
$id3 = $_POST["cat_2"];
$query2 = "select * from faqs_treeview where high_level=$id3";
$result2 = mysql_query($query2);
$form_visible = "visible";
}
else
{
$form_visible = "hidden";
}
?>
<td>
<form method="POST" name="myform3" id="t3" style="visibility: <?= $form_visible ?>">
<select name="cat_3" >
<option>Selecione a sub-sub-categoria</option>
<?php
while($row2 = mysql_fetch_array($result2))
{
$id = $row2["id_category"];
$name = $row2["category_title"];
echo "<option value='$id'>".$name."</option>";
}
?>
</select>
<input type="submit" name="submit3" onclick="closeall();">
<input type="hidden" name="cat_1" value="<?= $_POST["cat_1"]?>">
<input type="hidden" name="cat_2" value="<?= $_POST["cat_2"]?>">
</form>
</td>
</tr>
</table>
You could use AJAX for it. I see you use jQuery, so a simple get or post-request should do the trick.
Here: jQuery .get you will find some examples on how to do this. After the item has been added to the database you can use jQuery to add it to your select-box.
I am just building a simple search page in PHP. I need to know how can i keep the selecte value of the drop down list upon form submission. Currently, the value resets to the first index.
Can I do this via PHP without using client-side script?
Here is the code:
<?php
mysql_connect('localhost','root','');
mysql_select_db('hotel');
$sql = "SELECT * FROM customers";
$result = mysql_query($sql);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form method="get">
<select name="field" id="field">
<?php
/*if($field == 'Active')
'selected="selected"';
*/
while($rows = mysql_fetch_array($result))
echo '<option>'.$rows['customer_id'].'</option><br>';
?>
</select>
<?php
if (isset($_GET['Search']) && $_GET['action']=='search')
{
$sql="SELECT * FROM customers WHERE customer_id=".$_GET['field'];
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
echo '<br>Customer Name: '.$row['customer_name'].'<br>';
echo 'Email Address: '.$row['Email_Addr'].'<br>';
echo 'Contact No: '.$row['Contact_No'].'<br>';
}
?>
<input type="hidden" name="action" value="search" />
<br><input type="submit" value="search" name="Search" onclick="" />
</form>
</body>
</html>
usually like this.
echo '<option';
if ($_GET['field'] == $rows['customer_id']) echo " selected";
echo '>'.$rows['customer_id'].'</option>';
And please don't use the mysql_* functions to write new code, especially when you are learning. The mysql_* functions are in the process of becoming deprecated, they will be removed in future versions of PHP. Use mysqli_* or PDO objects instead.
You can check if the get value is the same than the select value :
while($rows = mysql_fetch_array($result))
echo '<option value="'.$rows['customer_id'].'" '.($rows['customer_id'] == $_GET['field']?'selected="selected"':'').'>'.$rows['customer_id'].'</option><br>';
When printing the select options, you can check for the value and set any mathing option to selected, maybe like this:
while($rows = mysql_fetch_array($result)){
if(!empty($_GET['field']) && $_GET['field'] == $rows['customer_id']){
echo '<option selected="selected">'.$rows['customer_id'].'</option><br>';
}
else {
echo '<option>'.$rows['customer_id'].'</option><br>';
}
}
I'm looking to make the the suggestions given by the Dictionary API to become links that can be queried or that are inserted directly in to the text are field before they are searched. I'm looking to achieve some kind of query expansion in any case.
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Search Attempt</title>
</head>
<body>
<form method="POST" action='AllinOneMonstaaa.php'>
<label for="query">Query</label><br/>
<input name="query" type="text" size="60" maxlength="60" value="" /><br /><br />
<select name ="agg">
<option value="Aggregated">Aggregated</option>
<option value="Non-Aggregated">Non-Aggregated</option>
<option value="Bing">Bing</option>
<option value="Blekko">Blekko</option>
<option value="Faroo">Faroo</option>
</select>
<input name="bt_search" type="submit" value="Search" /> </form>
<h2> Results </h2>
</body>
</html>
<?php
if ($_POST['query'])
{
$query = urlencode($_POST['query']);
$s_count = 0;
$ss_count = 0;
$query = 'http://www.dictionaryapi.com/api/v1/references/collegiate/xml/'.$query.'?
$xml = new SimpleXMLIterator(file_get_contents($query));
foreach ($xml -> suggestion as $suggestion[$s_count])
{
$s_count++;
}
if ($s_count > 1)
{
echo ('<h4>Did you mean?</h4>');
while ($ss_count <=$s_count)
{
echo ($suggestion[$ss_count].'<br>');
$ss_count++;
}
}
}
'Quick bug, there's a missing ' on the $query line.
echo '<a href="whatever">'. $suggestion[$ss_count].'<br>';
Edit: This will send them to Dictionary.com with the word to look up there. Outside that I have no idea how else you would formulate the url.
echo '<a href="http://dictionary.reference.com/browse/"' . $suggestion[$ss_count] . '" target="_new">' . $suggestion[$ss_count] . '<br>';
I have 3 files
1) show_createtable.html
2) do_showfielddef.php
3) do_showtble.php
1) First file is for creating a new table for a data base, it is a fom with 2 inputs, Table Name and Number of Fields. THIS WORKS FINE!
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<h1>Step 1: Name and Number</h1>
<form method="post" action="do_showfielddef.php" />
<p><strong>Table Name:</strong><br />
<input type="text" name="table_name" size="30" /></p>
<p><strong>Number of fields:</strong><br />
<input type="text" name="num_fields" size="30" /></p>
<p><input type="submit" name="submit" value="go to step2" /></p>
</form>
</body>
</html>
2) this script validates fields and createa another form to enter all the table rows.
This for also WORKS FINE!
<?php
//validate important input
if ((!$_POST[table_name]) || (!$_POST[num_fields])) {
header( "location: show_createtable.html");
exit;
}
//begin creating form for display
$form_block = "
<form action=\"do_createtable.php\" method=\"post\">
<input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\">
<table cellspacing=\"5\" cellpadding=\"5\">
<tr>
<th>Field Name</th><th>Field Type</th><th>Table Length</th>
</tr>";
//count from 0 until you reach the number fo fields
for ($i = 0; $i <$_POST[num_fields]; $i++) {
$form_block .="
<tr>
<td align=center><input type=\"texr\" name=\"field name[]\"
size=\"30\"></td>
<td align=center>
<select name=\"field_type[]\">
<option value=\"char\">char</option>
<option value=\"date\">date</option>
<option value=\"float\">float</option>
<option value=\"int\">int</option>
<option value=\"text\">text</option>
<option value=\"varchar\">varchar</option>
</select>
</td>
<td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\">
</td>
</tr>";
}
//finish up the form
$form_block .= "
<tr>
<td align=center colspan=3><input type =\"submit\" value=\"create table\">
</td>
</tr>
</table>
</form>";
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Create a database table: Step 2</title>
</head>
<body>
<h1>defnie fields for <? echo "$_POST[table_name]"; ?>
</h1>
<? echo "$form_block"; ?>
</body>
</html>
Problem is here
3) this form creates the tables and enteres them into the database.
I am getting an error on line 37 "Parse error: syntax error, unexpected $end in /home/admin/domains/domaina.com.au/public_html/do_createtable.php on line 37"
<?
$db_name = "testDB";
$connection = #mysql_connect("localhost", "admin_user", "pass")
or die(mysql_error());
$db = #mysql_select_db($db_name, $connection)
or die(mysql_error());
$sql = "CREATE TABLE $_POST[table_name](";
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
if ($_POST[field_length][$i] !="") {
$sql .=" (".$_POST[field_length][$i]."),";
} else {
$sql .=",";
}
$sql = substr($sql, 0, -1);
$sql .= ")";
$result = mysql_query($sql, $connection) or die(mysql_error());
if ($result) {
$msg = "<p>" .$_POST[table_name]." has been created!</p>";
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Create A Database Table: Step 3</title>
</head>
<body>
<h1>Adding table to <? echo "$db_name"; ?>...</h1>
<? echo "$msg"; ?>
</body>
</html>
$result = mysql_query($sql, $connection) or die(mysql_error());
if ($result) {
$msg = "<p>" .$_POST[table_name]." has been created!</p>";
}
you missing a } in your last if statement, and your for loop is missing a } too
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
if ($_POST[field_length][$i] !="") {
$sql .=" (".$_POST[field_length][$i]."),";
} else {
$sql .=",";
}
}
This error message means that a control structure block isn’t closed properly. In your case the closing } of some of your control structures like the for loop or the last if are missing.
You should use proper indentation and an editor that highlights bracket pairs to have a visual aid to avoid such errors.
You must close the for expression block:
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
// ...
}
Your for loop is not terminated. You are missing a }
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
}
And as pointed by others there is also a missing } for the last if statement:
if ($result) {
$msg = "< p>" .$_POST[table_name]." has been created!< /p>";
}
in your php.ini (php configuration) change :
short_open_tag = Off
you opened php tag shortly at line 1
just find and replace all <? with <?php
Stylistic tip: Use HEREDOCs to assign blocks of text to a variable, instead of the hideous multi-line-with-tons-of-escaping-backslashes constructs you're using. They're far easier to read and less error prone if/when you happen to forget a \ somewhere and break the script with a parse error.
I just solved this error, after checking my code, I had no open tags/braces.
For me, I got this error when moving to a amazon server.
It turns out I needed to enable short_open_tag = On in my php.ini.
This solved this error for me.