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Th3Boss

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Posts posted by Th3Boss

    insert different lines from <textarea> to different mysql rows

    in PHP Coding Help

    Posted

    Flip the parameter order of both array_filter() function calls.  That should fix all of the warnings.

     

    E.x.

    $textLines = array_filter('trim', $textLines); 

    to

    $textLines = array_filter($textLines, 'trim'); 

     

    nbst

     

    Yeah that fixed the errors.

    echo $query looks like this:

    INSERT INTO Table2 (`ID`,`Name`) VALUES ('50', 'Name1'), ('50', 'Name2'), ('50', 'Name3')

    Though it doesnt insert it into the table. It does insert the 50 into Table1 though from.

    $query = "INSERT INTO table1 (ID) VALUES (NULL)";

    insert different lines from <textarea> to different mysql rows

    in PHP Coding Help

    Posted · Edited by Th3Boss

     

    Try something like this:

     

    <?php

//Convert text input into lines based upon newline
$textLines = explode("\n", $_POST['TextArea']);
//Trim all the values
$textLines = array_map('trim', $textLines);
//Remove empty line
$textLines = array_filter('trim', $textLines);
//Escape input for query
$textLines = array_filter('mysql_real_escape_string', $textLines);

//Create parent record in table 1
$query = "INSERT INTO table1 (field1) VALUES ('value1')";
$result = mysql_query($query) or die(mysql_error());
$table1ID = mysql_insert_id();

//Process the lines into separate values for INSERT query
$insertValues = array();
foreach($textLines as $line)
{
    $insertValues[] = "('{$table1ID}', '{$line}')";
}

//Create INSERT query
$query = "INSERT INTO table2 (ID, Name) VALUES " . implode(', ', $insertValues);

?>

     

    Warning: array_filter() expects parameter 1 to be array, string given in on line 60

     

    $textLines = array_filter('trim', $textLines);
 

     

    Warning: array_filter() expects parameter 1 to be array, string given on line 62

    $textLines = array_filter('mysql_real_escape_string', $textLines);
 

     

    Warning: Invalid argument supplied for foreach() on line 71

    foreach($textLines as $line)
 

     

    And should field1 be a new column for the parent record or should that be the auto increment ID from table 1?

     

     

    Try something like this:

    //Create parent record in table 1
$query = "INSERT INTO table1 (field1) VALUES ('value1')";
$result = mysql_query($query) or die(mysql_error());
$table1ID = mysql_insert_id();
 

     

    Like:

    //Create parent record in table 1
$query = "INSERT INTO table1 (ID) VALUES (NULL)";
$result = mysql_query($query) or die(mysql_error());
$table1ID = mysql_insert_id();
 

    insert different lines from <textarea> to different mysql rows

    in PHP Coding Help

    Posted · Edited by Th3Boss

    oh I just noticed I have that wrong.

     

    The ID needs to be the same as the ID from another table that is auto increment, the structure is this.

    It inserts the ID row # for Table 1 as auto increment and then inserts the ID from Table 1 into Table 2 with the data from the textarea.

     

     

    From Textarea:

    Info1

    Info2

    Info3

    Info4

    Info5

     

    Table 1:

    ID

    1

    2

    3

    4

     

     

    Into Table2:

    ID | Name

    4  | Info1

    4  | Info2

    4  | Info3

    4  | Info4

    4  | Info5

    insert different lines from <textarea> to different mysql rows

    in PHP Coding Help

    Posted · Edited by Th3Boss

    A. You need to normalize your data.

    B. what debugging have you done? Echo the query, run it in MySQL/phpMyAdmin, etc?

     

    But ignore B until after you do A.

     

     

    Edit: That's assuming your structure looks like that. You said rows, but your code doesn't look like it's trying to do rows. Do you have many COLUMNS with those names, or you want many rows? (Hint: rows is right.)

     

    What do you mean normalize the data?

     

    the structure I want is this

     

    From Textarea:

    Info1

    Info2

    Info3

    Info4

    Info5

     

    Into Table:

    ID | Name

    1  | Info1

    1  | Info2

    1  | Info3

    1  | Info4

    1  | Info5

    insert different lines from <textarea> to different mysql rows

    in PHP Coding Help

    Posted

    Well I replied to an old thread but as it is old it doesnt seem to be getting any activity as it is marked answered.

     

    The thread is: http://forums.phpfreaks.com/topic/228584-insert-different-lines-from-textarea-to-different-mysql-rows/

     

    This doesnt work. I have tried it so many different ways.

     

    I have a textarea that has data like

    Info1

    Info2

    Info3

    Info4

    Info5

     

    I have a table that has an auto increment ID and a Name column and I want each line from the textarea in its own row inserted in the Name column.

     

    The only reply to the thread says to try:

     

     

    $lines = array_map('mysql_real_escape_string', explode("\n", $_POST['textarea']));
$values = "VALUES ('" . implode("'), ('", $lines) . "')";

$query = "INSERT INTO table_name (column_name) $values";
 

    Which didnt work. Ive tried a bunch of different things and cant get this to work.

     

    Here is my last attempt of trying to get the above to work:

     

     

    $lines = array_map('mysql_real_escape_string', explode("\n", $_POST['TextArea']));
$values = implode("'), ('", $lines);

$query = "INSERT INTO Table (`ID`,`Name`) VALUES (NULL, '$values')";
 

    insert different lines from <textarea> to different mysql rows

    in PHP Coding Help

    Posted

     

    One way:

     

    $lines = array_map('mysql_real_escape_string', explode("\n", $_POST['textarea']));
$values = "VALUES ('" . implode("'), ('", $lines) . "')";

$query = "INSERT INTO table_name (column_name) $values";

     

     

    This doesnt work. I have tried it so many different ways.

     

    I have a textarea that has data like

    Info1

    Info2

    Info3

    Info4

    Info5

     

    I have a table that has an auto increment ID and a Name column and I want each line from the textarea in its own row inserted in the Name column.

    I tried like posted above and so many other ways. Here is my last attempt.

     

    $lines = array_map('mysql_real_escape_string', explode("\n", $_POST['TextArea']));

    $values = implode("'), ('", $lines);

     

    $query = "INSERT INTO Table (`ID`,`Name`) VALUES (NULL, '$values')";

    help needed

    in PHP Coding Help

    Posted

    Why don't you select those where slot = 'high' in the query and set LIMIT 8?

     

    Right now in the query it selects all of that data i need from the database and now I need to set all of that data as the correct variables.

    help needed

    in PHP Coding Help

    Posted · Edited by Th3Boss

    Pretty sure im doing this completely wrong.

     

     

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

    if($row['slot'] == 'high') {
        $i=1;
        while($i<=
        {
        $name[$i]  .= $row['name']."" ;
        $typeID[$i]  .= $row['typeID']."" ;
        $i++;
        }
    } else {      

    }
}

     

    There will always be 0 to 8 items that are 'high' in the database $row['slot'].

    I want all of the 'name' and 'typeID' to be as:

     

    $name1
    $typeID1

    $name2
    $typeID2

    $name3
    $typeID3

     

    and so on up to how ever many there is, but no more than 8

     

    How do I do something like this? (Sorry if that is a terrible explanation)

    syntax error help

    in MySQL Help

    Posted

    This is the error I keep getting and I cant figure out where the error is.


     

     

    SQL query:

    -- -- Dumping data for table `ships` -- INSERT INTO `ships` (`Ship_Name`, `typeID`, `high`, `mid`, `low`, `rig`, `drone`) VALUES ('Bantam', 582, 3, 4, 2, 3, 5, ''), ('Condor', 583, 4, 4, 2, 3, 0, ''), ('Griffin', 584, 2, 5, 2, 3, 5, ''), ('slasher', 585, 4, 4, 2, 3, 0, ''), ('Probe', 586, 3, 4, 3, 3, 5, ''), ('Rifter', 587, 4, 3, 3, 3, 0, ''), ('Executioner', 589, 4, 3, 3, 3, 0, ''), ('Inquisitor', 590, 3, 2, 4, 3, 5, ''), ('Tormentor', 591, 3, 3, 4, 3, 5, ''), ('Navitas', 592, 3, 3, 3, 3, 5, ''), ('Tristan', 593, 3, 3, 3, 3, 5, ''), ('Incursus', 594, 3, 3, 4, 3, 5, ''), ('Punisher', 597, 4, 2, 4, 3, 0, ''), ('Breacher', 598, 3, 4, 3, 3, 5, ''), ('Burst', 599, 3, 3, 3, 3, 5, ''), ('Kestrel', 602, 4, 4, 2, 3, 0, ''), ('Merlin', 603, 3, 4, 3, 3, 0, ''), ('Heron', 605, 3, 5, 2, 3, 5, ''), ('Imicus', 607, 3, 4, 3, 3, 5, ''), ('Atron', 608, 4, 3, 3, 3, 0, ''), ('Maulus', 609, 2, 4, 3, 3, 5, ''), ('Crucifier', 2161, 2, 4, 3, 3, 5, ''), ('Echelon', 3532, 0, 1, 0, 0, 0, ''), ('Vigil', 3766,[...]

    MySQL said: b_help.png

    #1064 - 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 293

     

     
    Anyone see what is wrong with it?

    line break help

    in Javascript Help

    Posted

    Im trying to get it to place a line break after each output

     

        function generateTemplate() {
        output = '$' + templateName + ';' + "<br>";
        output += '$ship=' + fit.shipName + ';' + "<br>";
        output += '$shipTypeID=' + fit.shipTypeID + ';' + "<br>";
        output += '$fitName=' + fit.name + ';' + "<br>";
        output += '$fitID=' + fit.fitID + ';' + "<br>";
    }
 

     

    output is displayed in a textarea

     

    How do I do this?

    Javascript Help

    in Javascript Help

    Posted

    well you can start by explaining what it is you are tying to accomplish...

     

    On click of class="Info" it uses $DataTypeID for data-typeid from page 1 and does showItemInfo using that data form data-typeid as typeID.

    Javascript Help

    in Javascript Help

    Posted

    Never used javascript before so trying to figure this out.

     

    Page 1:

    <?php
$DataTypeID="599";
include("Page2.php");
?>

 

     

    Page 2:

    <head>
<script type="text/javascript">
$(function() {
$('.Fitting .Info').on('click', showItemInfo);
});

function showItemInfo() {
var typeID = $(this).attr('data-typeid');
CCPEVE.showInfo(typeID);
}
</script>
</head>
<body>
<div class="Fitting">
<div class="Info" data-typeid == <?$DataTypeID?> title="View">INFO</div>
</div>
</body>
 

     

    Pretty sure this is wrong

     

     

    data-typeid == <?$DataTypeID?>
 

     

    How do I get this to work?

    if statements

    in PHP Coding Help

    Posted

     

    You shouldn't jump in and out of PHP tags like that, it makes for messy code.

     

    Try this:

     

    <?PHP

  if($highname1 == 'open') {
    $output = '<div class="module high open">
                 <img src="Icon_hi_slot.png" width="46" height="46">
               </div>';
  } else if(strlen(trim($highname1))) {      
    $output = '<div class="module high inactive">
               </div>';
  } else {      
    $output = '<div class="module high" data-typeid=$high1typeID data-name=$high1name>
                 <img src="http://image.website.com/Type/'. $high1typeID .'_64.png">
               </div>';
  }
 
  echo $output;
 
?>

     

     

    That worked though they were switched around. Works as wanted like this:

     

     

    <?PHP

  if($high2name == 'open') {
    $output = '<div class="module high open">
                 <img src="Icon_hi_slot.png" width="46" height="46">
               </div>';
  } else if(strlen(trim($high2name))) {      
    $output = '<div class="module high" data-typeid=$high2typeID data-name=$high2name>
                 <img src="http://image.website.com/Type/'. $high2typeID .'_64.png">
               </div>';
  } else {      
    $output = '<div class="module high inactive">
               </div>';
  }
 
  echo $output;
 
?>
 

     

    Thanks

    if statements

    in PHP Coding Help

    Posted

    ok and it seems this does the same thing but as an if statement

     

    <?php if ($high1name == "open") { ?>

<div class="module high open">
    <img src="Icon_hi_slot.png" width="46" height="46">
</div>

<?php } elseif ($high1name == "") { ?>

    <div class="module high inactive" >

    </div>

<?php } else { ?>

<div class="module high" data-typeid=$high1typeID data-name=$high1name>
    <img src="http://image.website.com/Type/<?php print($high1typeID);?>_64.png">
    
</div>


<? } ?>
 
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