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klinmy

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Posts posted by klinmy

    Display selected variable(s)

    in PHP Coding Help

    Posted

    Hi, I'm having problem to display the selected data in the table. The codes are as follow,

     

    
// To print table header of Date and Time
echo "   <table width='120%' border='1' cellspacing='0' cellpadding='0'>";
echo "<tr>
<th  bgcolor='#CCCCCC'>Date</th>
<th  bgcolor='#CCCCCC'>Time</th>";


// To print table header for the selected variables ($var)
for ($i=0; $i<$c; $i++)
{

$var=$v[$i];  // the selected variable. eg: $v[0] is Height;  $v[1] is Weight;

echo "<th  bgcolor='#CCCCCC'>$var</th>";


} // <- no problem with this

echo "</tr>";


while ($row = mysql_fetch_array($result)) 
{


echo '<tr><td align=center>' . $row['Date'] . '</td>';
echo '<td align=center>' . $row['Time'] . '</td>';
echo '<td align=center>' . $row[$var] . '</td>';

echo "</tr>"; 


} //end while

     

    And it's an example of the output:

     

    Date    Time    Height            Weight

    12 feb    10.00am    40 <-(it's the weight)

    15 feb    11.00am    60

    16 feb    12.00am    70

    17 feb    11.00am    20

    18 feb    01.00am    40

     

    It displayed only the weight ($v[1]) values under Height.

    And when I tried putting the while loop inside the for loop, it displayed only the height values.

     

    What it should be done in order to print both selected variables (height & weight) together with the date and time into the table?

     

    thanks in advance.

    Display selected data in a table

    in PHP Coding Help

    Posted

    Hi, I would like to display the selected data in a table.

     

    
echo "<tr><td>";
echo "Data1;
echo "</td></tr>";

while ($row=mysql_fetch_assoc($result))
{
echo "<tr><td>";
echo "$row["data1"];
echo "</td></tr>";
}

-----some other codes-------

echo "<tr><td>";
echo "Data3;
echo "</td></tr>";

while ($row2=mysql_fetch_assoc($result))
{
echo "<tr><td>";
echo "$row2["data3"];
echo "</td></tr>";
}

     

    The result I got is something like ths,

     

    Data1

    12

    4

    1

    2

    3

    Data3

    5

    4

    2

    6

    4

     

    Is there a way that i can make the table looks like this?

     

    Data1    Data2

    12            5

    4              4

    1              2

    2              6

    3              4

     

    Thanks in advance

     

    Array

    in PHP Coding Help

    Posted

    Hi, i would need this type of array data e.g. $new=array(12,23,9,2) to be used to plot the graph (axis-y), $plot = new BarPlot($new);

     

    however, if the array data is retrieved from the database, how can i map the retrieved data (eg: temperature = 21, 20, 19, 15) into $new = array (21,20,19,15)?

    while ($row = mysql_fetch_assoc($result)) 
	{	

	$temp = $row["Temperature"];


	}

     

    it retrieved the data as (21201915), what should i do to map this into $new=array (21,20,19,15).

     

    thanks in advance.

    Can't update record

    in PHP Coding Help

    Posted

    thanks blueman378, but it shows the same error after changed it to

     

    $result=mysql_query($query) or die(mysql_error());

     

    "You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1"

     

    <html>
<head>
<title>City</title>

</head>

<body>

<form name="frm" form action="" method="post" >
<?PHP

function check_mysql()
{
if(mysql_errno()>0)
{
	die("MySQL error". mysql_errno() . ":" . mysql_error());
}
}

$db=mysql_connect("localhost","root","");
if(!$db)
{
die("Failed to open connection to MySQL server.");
}

mysql_select_db("TPS");
check_mysql();

mysql_select_db("TPS");
check_mysql();

if(!isset($_POST['ID']))
{
$ID=0;
}

if(isset($_POST['save']))
{  	
$CityName=$_POST["CityName"];	
$errors=0;	
if($errors==0)
{				
	$query="INSERT INTO city" . "(  CityName) VALUES ('$CityName')";
	$result=mysql_query($query);
	check_mysql();
			$ID=mysql_insert_id();
}
}

elseif(isset($_POST['update']))
{
$error=0;	
	if($error==0)
	{					

$query="UPDATE City SET CityName='$CityName'" . "WHERE ID=$ID ";
$result=mysql_query($query) or die(mysql_error());

$message="<script>"
	    		." alert('Record Updated');"

				."</script>"; 

}	
}
?>
</div>

<div id="Layer1" style="position:absolute; left:238px; top:105px; width:662px; height:131px; z-index:4">
  <p>City Name : 
    <input type="text" name="CityName" <?php echo "VALUE=\"$CityName\""?> >
  </p>
  <p><span class="style16">.</span>  </p>
	  
    
      <input name="update" type="submit" id="update" value="Update "style=" width:8em;" />
         <input name="save" type="submit" id="save2"  value="Save " style=" width:7em;" /   >
     <input type="HIDDEN" name="ID" <?php echo "VALUE=\"$ID\""?>>
    <?PHP
if(isset($message))
{
echo "<BR><BR>$message";
}
?> 
    
</body>
</html>

     

    this is the simplify code which is from A-Z just in case there's other possibility that causes the error. i've tried every way i know to solve the problem, killed my whole day.

     

    thanks in advance.

    Can't update record

    in PHP Coding Help

    Posted

    Tried this as well, yet the record is not updated.

     

    elseif(isset($_POST['update']))

    {

    $error=0;

    $ID=$_POST["ID"];

     

    if($error==0)

    {

     

     

    $query="UPDATE City SET CityName='$CityName'" . "WHERE ID=$ID";

    $result=mysql_query($query);

    check_mysql();

    $message="<script>"

        ." alert('Updated');"

     

    ."</script>";

     

    }

     

    }

     

    What seems to be wrong?

    Can't update record

    in PHP Coding Help

    Posted

    Thanks blueman378, my mistake. But after changing it, it shows this error when i click the update button.

     

    "MySQL error1064:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1"

     

    Any possible answer for this? Dun find anything wrong with the update function.

     

    Thanks.

    Can't update record

    in PHP Coding Help

    Posted

    I've been trying it for the whole day, but i just can't figure what's the problem with the code..

     

    It's a form to record the city name, where the table (City) contains only 2 fields which are ID (auto_increment) and CityName.

     

    Below is the code for the form. There'r no problem on 'search', 'save', and  'delete', but the conditions for the functions have to be set as, for example $query="DELETE FROM City where CityName='$CityName' rather than $query="DELETE FROM City where ID='$ID'

     

    So there's where the "update" function comes.

     

    if(isset($_POST['ID']))
{
$ID=0;
}

if(isset($_POST['save']))
{  


$CityName=$_POST["CityName"];


$errors=0;


if($errors==0)
{			


	$query="INSERT INTO city" . "(  CityName) VALUES ('$CityName')";
	$result=mysql_query($query);
	check_mysql();
	$message = "<script>"
	    		." alert('Record Saved');"

				."</script>";
	$CityName="";
	$ID=mysql_insert_id();
}
}

elseif(isset($_POST['search']))
{

$ID=0;
$CityName=$_POST["CityName"];


$query5="SELECT  CityName FROM City ". "WHERE CityName= '$CityName' AND ID > $ID";
$result5=mysql_query($query5);
check_mysql();
$row5=mysql_fetch_row($result5);
check_mysql();

if($row5>0)
{

	$CityName=$row5[0];


	$message = "<script>"
	    		." alert('Found $IDfound');"

				."</script>";
				$IDfound='$IDfound';
}
else
{
$message = "<script>"
	    		." alert('No Record Found');"

				."</script>";
	//$CityName="";

}
}
elseif(isset($_POST['update']))
{
$query="UPDATE City SET CityName='$CityName'" . "WHERE ID=$ID";
$result=mysql_query($query);
check_mysql();
$message="<script>"
	    		." alert('Record Update');"

				."</script>"; 

}

elseif(isset($_POST['delete']))
{
$CityName=$_POST["CityName"];


$error1=0;


	if($error1==0)
	{

		$query="DELETE FROM City where CityName='$CityName'" ;
		$result=mysql_query($query);
		check_mysql();



		$message="<script>"
	    		." alert('Record deleted');"

				."</script>"; 
	$CityName="";
	}



}

     

    thanks in advance. I tried looking at other codes too, but still can't able to find the solution for it..

    passing data of radio button

    in PHP Coding Help

    Posted

    radio.php
    [code]<form method="post" action="nextpage.php">
    <?

    $query=mysql_query("SELECT name FROM staff WHERE id='1122' ");
    while ($row = mysql_fetch_array($query))
    {
        echo "$row[name]";
        $name=$row[name];
        
    }

    ?>
    <input type="radio" name="answer" value="<? $_POST['name'] ?>">
    <input type="radio" name="answer" value="B">

    <input type="submit" name="Submit" value="Submit">
    </form>[/code]


    nextpage.php
    [code]<? echo '1122 is  '.$_POST['name']!; ?>[/code]

    when i clicked in the 1st radio button, it displayed "[!--coloro:#3333FF--][span style=\"color:#3333FF\"][!--/coloro--]1122 is ! [!--colorc--][/span][!--/colorc--]"
    why cant it display the name as it's shown in radio.php? thanks

    update all records at once?

    in PHP Coding Help

    Posted

    i'm doing a form which the admin can enter a particular date for the system's users. Different users may have different dates which will be deducted by the current date in order to get the duration.

    my question is, can i make a button which can update all the duration of the system's users in the database, so that the admin doesn't need to update the duration one by one when it enters a new year?

    thanks

    auto_increment id

    in PHP Coding Help

    Posted

    it's now working fine after changing to ths..

    [code]if(isset($save))
    {        
        
        
        $query="INSERT INTO apply " . "(  name, sex) VALUES ( '$name', '$sex')";
        $result=mysql_query($query);
        check_mysql();
        $id=mysql_insert_id();
            
        header("Location: FormB.php?id=$id&name=$_POST[name]&sex=$_POST[sex]");exit;
        
    }[/code]

    thanks guys

    auto_increment id

    in PHP Coding Help

    Posted

    there isn't any GET for id in Form A. or is there??

    i'm trying to get the value of id to Form B. i've tried using both $_POST[id] and $_POST['id']. but it just made no different..name and sex can stil get the values as entered in form A

    auto_increment id

    in PHP Coding Help

    Posted

    it's the code i tried on form A

    i now change the next button to save.

    [code]mysql_select_db("onlineleave");
    check_mysql();

    if(!isset($id))
        $id=0;


    if(isset($save)){    
        
        $query="INSERT INTO apply " . "(name, sex) VALUES ( '$name', '$sex')";
        $result=mysql_query($query);
        check_mysql();
        $id=mysql_insert_id();
                    echo "id = $id";
    }[/code]
    the message showed the correct id as in database, no problem it seems.

    but when i bring the info to Form B,

    [code]if(isset($save))
    {        
        
        
        $query="INSERT INTO apply " . "(  name, sex) VALUES ( '$name', '$sex')";
        $result=mysql_query($query);
        check_mysql();
        $id=mysql_insert_id();
            
        header("Location: FormB.php?id=$_POST[id]&name=$_POST[name]&sex=$_POST[sex]");exit;
        
    }[/code]

    and placing this in Form B,

    [code]<?php echo '<pre>'.print_r($_GET,true).'</pre>'; ?>[/code]

    it came out like this,

    Array
    (
    [name] => john
    [id] => 0
    [sex] => male

    )

    why is it cant get the value id (auto_increment)?

    thanks alot!

    auto_increment id

    in PHP Coding Help

    Posted

    thanks txmedic03, but all the other values (name and sex) are working fine except for id only.

    i tried to close my Form B. by running only Form A, nomatter how many times i enter data, the id still remains '0'..

    auto_increment id

    in PHP Coding Help

    Posted

    when i fill up form A, the information will be displayed in the next page (form B) when i click the "next" button.

    [!--coloro:#3333FF--][span style=\"color:#3333FF\"][!--/coloro--]example data that i entered in Form A in the 1st time,
    id = (auto_increment)
    name=john
    sex = male

    example data which is displayed in Form B in the 1st time,
    id=0 (supposely 221)
    name=john
    sex = male[!--colorc--][/span][!--/colorc--]

    [!--coloro:#993399--][span style=\"color:#993399\"][!--/coloro--]example data that i entered in Form A in the 2nd time,
    id = (auto_increment)
    name=selina
    sex = female

    example data which is displayed in Form B in the 2nd time,
    id=221 (supposely 222)
    name=selina
    sex = female[!--colorc--][/span][!--/colorc--]


    the problem is when the first time i fill up the form, the auto_increment id will not be displayed in form B, it will only be displayed with the first id when i fill up for the second time.

    below is parts of my script in form A,

    [code]$db=mysql_connect("localhost","ky","pw");
    if(!$db)
    {
        die("Failed to open connection to MySQL server.");
    }

    mysql_select_db("onlineleave");
    check_mysql();

    if(!isset($id))
    {
        $id=0;
    }

    elseif(isset($next))
    {
            
        if($errors==0)
        {        

        
        check_mysql();
        $id=mysql_insert_id();
        
        header("Location: applyBa.php?name=$_POST[name]&id=$_POST[id]&sex=$_POST[sex]"); exit;
        
        }    
    }

    <input type="hidden" name="id" <?php echo "VALUE=\"$id\""?>>[/code]

    Form B

    [code]<?php echo $_GET[name] ?> is <?php echo $_GET[sex] ?> with ID <?php echo $_GET[id] ?>.

    if(isset($save))
    {
        
        $query="INSERT INTO tableB " . "( name, sex) VALUES ( '$name, '$sex)";
        $result=mysql_query($query);
        check_mysql();
        $id=mysql_insert_id();[/code]


    thanks
    the id value in the database is correct, just the display in form B has problem.
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