Jump to content

Getting mysql data into an array


mdmartiny

Recommended Posts

I am in the process of writing code for a dynamic form. I have the part of the form written for information to be entered. I am now trying to write code for the modification of the  information in the table.

 

I know that I want to get the table data my doing a MySql Select and putting that information into an array. I am not very familiar with arrays and have spent some time looking for a solution online with no solution. When I print_r the results I get what I want but when I do anything else I get Array. What I am actually looking to do is make it so that the table data is in another variable

 

this is the section of code that I am working with.

$mod_form = "<form name='insert_table' id='insert_table' action='ad_add.php' method='post' enctype='multipart/form-data'>
<fieldset>
<legend>table information</legend>
<input name='table' TYPE='hidden' VALUE='" . $table . "' />";

$mod_sql = "SELECT * FROM $table WHERE id = '$id'";
$mod_sql_result = mysql_query($mod_sql);

$allowed_ext = array('jpg', 'jpeg', 'png', 'gif');
$input_array = array();

while ($row = mysql_fetch_array($mod_sql_result, MYSQL_NUM)) {
    $input_array[] = $row;
}

foreach ($input_array as $input => $data) {
    print_r($data);
}

while ($i < mysql_num_fields($mod_sql_result)) {


    $header = str_replace("_", " ", (mysql_field_name($mod_sql_result, $i)));
    $name = mysql_field_name($mod_sql_result, $i);

    (mysql_field_name($mod_sql_result, $i) == 'image' ? $mod_form .= "<p>" . $header . ": <input name='" . $name . "' id='" . $name . "' type='file' /></p>" : $mod_form .= "<p>" . $header . ": <input name='" . $name . "' id='" . $name . "' type='text' value='" . $input . "'/></p>");

    $i++;
}
$mod_form .= "
<p>
<input type='submit' name='submit_mod' id='submit_mod' />
</p>
</fieldset>
</form>
";

echo $mod_form;

Link to comment
Share on other sites

This is just the test database that I am working with. Once this is done I will not know what the column names are going to be. The person that this is for will make all of the fields and names themselves. I am just writing the php code

Link to comment
Share on other sites

Here you go

<?php
//get sql
$Sql = "SELECT * FROM {$table} WHERE id = '{$id}'";
//run query
$Query = mysql_query($Sql);
//allowed ext
$allowed_ext = array('jpg', 'jpeg', 'png', 'gif');
//input array)
$input_array = array();
//$data 
$info = mysql_fetch_array($Query);
foreach ($info as $val) {
$input_array[] = $val;
}
//now start the form
$Form = "<form name=\"insert_table\" id=\"insert_table\" action=\"ad_add.php\" method=\"post\" enctype=\"multipart/form-data\">
<fieldset>
<legend> Table Information </legend>
<input name=\"table\" type=\"hidden\" value=\"{$table}\">";
//run foreach lop now
foreach ($input_array as $key => $val) {
$header = str_replace("_"," ",$key);
$Form .= ($key == 'image' ? "<p>{$header}: <input name=\"{$key}\" id=\"{$key}\" type=\"file\" /></p>" : "<p>{$header}: <input name=\"{$key}\" id=\"{$key}\" type=\"text\" value=\"{$val}\" /></p>" );
}
$Form .= "<p>
<input type=\"submit\" name=\"submit_mod\" id=\"submit_mod\" />
</p>
</fieldset>
</form>";
echo $Form;
?>

Link to comment
Share on other sites

try

<?php
//get sql
$Sql = "SELECT * FROM {$table} WHERE id = '{$id}'";
//run query
$Query = mysql_query($Sql);
//allowed ext
$allowed_ext = array('jpg', 'jpeg', 'png', 'gif');
//input array)
$input_array = array();
//$data 
$info = mysql_fetch_array($Query);
//now start the form
$Form = "<form name=\"insert_table\" id=\"insert_table\" action=\"ad_add.php\" method=\"post\" enctype=\"multipart/form-data\">
<fieldset>
<legend> Table Information </legend>
<input name=\"table\" type=\"hidden\" value=\"{$table}\">";
//run foreach lop now
foreach ($info as $key => $val) {
$header = str_replace("_"," ",$key);
$Form .= ($key == 'image' ? "<p>{$header}: <input name=\"{$key}\" id=\"{$key}\" type=\"file\" /></p>" : "<p>{$header}: <input name=\"{$key}\" id=\"{$key}\" type=\"text\" value=\"{$val}\" /></p>" );
}
$Form .= "<p>
<input type=\"submit\" name=\"submit_mod\" id=\"submit_mod\" />
</p>
</fieldset>
</form>";
echo $Form;
?>

Link to comment
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.