option value as part of form block

buchberg · February 15, 2014

Hello, and thank you in advance for looking at this. I am trying to write a piece of code for an input value using an option dropdown box. In my mind I am obviously in need of some assistance.

This is my set up:

Apache 2.21

PHP: 5.3.5

MySql: 5.6 (I think)

The following code has been working well for the mysql fetch array and display, but is outside my formblock. The problem that I am having is I would like it to be placed inside my form block.

Option Value code

// setting up fetch row for OWNER 19 //

$result_engineer= mysql_query("SELECT organization, l_name, f_name FROM $table_name WHERE type='municipality'") or die (mysql_error());

echo '<select name="owner[]">'; // Open your drop down box

while ($row_engineer = mysql_fetch_array($result_engineer)) {

echo '<option value>'.$row_engineer['organization'].', '.$row_engineer['l_name'].', '.$row_engineer['f_name'].'</option>';

}

echo '</select>';// Close your drop down box

My form block code starts as follows:

$form_block="

<P><b>Project Status:</b>

<br>

<br>

<P><small><small>Project Name:</small></small>

(it continues ad nauseum)

There seems to be an error once I place the Option value code inside the form block code. My suspicion is I am not using a proper <TAG>. I am new, but have been trying. Security is not much of an issue as this is on a local machine only.

Any assistance is greatly appreciated. Thank you and have a nice day.

Best regards

Ch0cu3r · February 15, 2014

Rather echo out the the <select>/</option> tags assign them to a variable

$engineerList = '<select name="owner[]">'; // Open your drop down box
 
while ($row_engineer = mysql_fetch_array($result_engineer)) {
   $engineerList .= '<option value>'.$row_engineer['organization'].', '.$row_engineer['l_name'].', '.$row_engineer['f_name'].'</option>';
}
 
$engineerList .= '</select>';// Close your drop down box

Now place $engineerList in the $form_block string where you want the select menu to appear. Example

$form_block="
<FORM METHOD=\"POST\"ACTION=\"do_add_design_XXXXXX.php\">
...
<P>Engineer: $engineerList</P> <!-- add engineer list -->
...
</FORM>";

Edited February 15, 2014 by Ch0cu3r

buchberg · February 16, 2014

Thank you so much. I will give it a shot. I really appreciate the input, and have a good day.

buchberg · March 1, 2014

Thank you. It worked awesome.

My value is showing up as an Array on my confirmation html page with command

echo $_POST[engineer];

However, it shows nothing in my phpmyadmin. I would expect Array to show up.

In using the suggested code ($engineerList = '<select name="owner[]">'; // Open your drop down box), I changed "ower[]" to "engineer[]".

.

As I am seeing in my confirmation page the value Array. I did use the following code to grab the values:

foreach ($_POST[engineer] as $key=>$v_eng){

$v_eng=implode(',',$_POST['engineer']);

}

It has worked on other variables except this one Is there a difference in the way the list is generated that doesn't allow it to POST in the same way?

Thank you in advance, and have a good day.

Ch0cu3r · March 2, 2014

Remove the [] in the name. You only need to have that in the name if you are naming other fields the with same name.

buchberg · March 4, 2014

Your help is very much appreciated, but this is frustrating me to no end. The mysql fetch side of $engineerList works great. However, I am still having issues with getting a $_POST[engineer] value recognized/saved in the form.

I implemented the following code as we discussed, and removed the [] in $engineerList ='<select name="engineer">'.

// setting up fetch row for OWNER 19 //

$result_engineer= mysql_query("SELECT organization, l_name, f_name FROM $table_name WHERE type='engineer'") or die (mysql_error());

$engineerList = '<select name="engineer"> '; // Open your drop down box

while ($row_engineer = mysql_fetch_array ($result_engineer)) {

$engineerList .= '<option value>'.$row_engineer['organization'].', '.$row_engineer['l_name'].', '.$row_engineer['f_name'].'</option>';

}

$engineerList .= '</select> ';// Close your drop down box

This displays all values in the MySQL table associated with engineers. But whatever engineer I select from this list, it isn't saved and doesn't have a value to INSERT into my table in MySQL.

To verify, I have tried echo $_POST[engineer]; and there is no value.

Am I looking for the wrong variable to be stored, does the code not recognize the $engineerList as an input?

Please let me know any suggestions that you have. Thank you and I know this is frustrating. I really appreciate your input, and have a good day.

Ch0cu3r · March 4, 2014

Can you post your code here. It sounds to me you are not including the engineer list within your form.

buchberg · March 5, 2014

The $engineerList is within the $form_block, but it is not recognized by the server side code. On the ser.ver side, I have error checks if the value of $_POST$engineerList = "".

Please let me know where I am going wrong. Thank you and I look forward to your reply.

<HTML>

<HEAD>

<TITLE>Project Bid Interface</TITLE>

</HEAD>

<BODY>

<br>

<div style="text-align: left;"><img

style=" width: 250px; height: 150px;" alt="This is family pic" src="family_pic.jpg"><big><big><big><big><span

style="font-weight:"> <small>This is my Work!!!<br>

<br>

<?

$db_name="XXXXX";

$table_name="yyyyyy";

$connection=mysql_connect("localhost", "root", "password") or die (mysql_error());

mysql_select_db("$db_name") or die (mysql_error());

// setting up fetch row for OWNER 19 //

$result_engineer= mysql_query("SELECT organization, l_name, f_name FROM $table_name WHERE type='engineer'") or die (mysql_error());

$engineerList = '<select name="engineer"> '; // Open your drop down box

while ($row_engineer = mysql_fetch_array ($result_engineer)) {

$engineerList .= '<option value>'.$row_engineer['organization'].', '.$row_engineer['l_name'].', '.$row_engineer['f_name'].'</option>';

}

$engineerList .= '</select>';// Close your drop down box

// line 28

$form_block="

<P><b>Project Status:</b>

<br>

<br>

<P><small><small>Project Name:</small></small>

</P>

</fieldset>

<P><b>Contact Information:</b>

<P><small><small>Rep Territory:</small></small>

<br>

<INPUT TYPE=\"radio\" NAME=\"rep\" Value=\"XX\"><small><small>Michael Buchberg</small></small><br>

<INPUT TYPE=\"radio\" NAME=\"rep\" Value=\"XY\"><small><small>Mike Krebs</small></small><br>

<INPUT TYPE=\"radio\" NAME=\"rep\" Value=\"XZ\"><small><small>Tom McNamara</small></small><br>

<P><small><small>Owner:</small></small><br>

<INPUT TYPE=\"radio\" NAME=\"owner\" Value=\"XX\"><small><small>Mike Krebs</small></small><br>

<INPUT TYPE=\"radio\" NAME=\"owner\" Value=\"YY\"><small><small>Tom McNamara</small></small><br>

<P><small><small>Engineer:</small></small>

<p>$engineerList</p>

</FIELDSET>

<br>

<P><small><small>Equipment:</small></small><br>

<INPUT TYPE=\"checkbox\" NAME=\"equipment[]\" Value=\"none \"><small><small>No Equipment</small></small>

<INPUT TYPE=\"checkbox\" NAME=\"equipment[]\" Value=\"odor\"><small><small>Odor Control</small></small>

<INPUT TYPE=\"checkbox\" NAME=\"equipment[]\" Value=\"pumps\"><small><small>Pumps</small></small>

<INPUT TYPE=\"checkbox\" NAME=\"equipment[]\" Value=\"service\"><small><small>Service</small></small>

<br>

<small><small>Odor Control</small></small>

<input type=\"text\" name=\"odor\" <input name=\"v_odor\" <input=\"\"

onkeyup=\"this.value=this.value.replace(/[^\d]/,'')\"><br>

<small><small>Pumps</small></small>

<input name=\"v_pumps\" <input=\"\"

onkeyup=\"this.value=this.value.replace(/[^\d]/,'')\"

type=\"text\"><br>

<small><small>Service</small></small>

<input name=\"v_service\" <input=\"\"

onkeyup=\"this.value=this.value.replace(/[^\d]/,'')\"

type=\"text\"><br><br>

</fieldset>

<P><b>Comments & Follow Up:</b>

<br>

</FORM>";

if ($_POST[op]!="ds") {

// 18 they need to see the form

echo "$form_block";

} else if ($_POST[op] == "ds") {

}

?>

</body>

</div>

</body>

</html>

server side (portion relating to inserting into table)

$db_name="XXXXX";

$table_name="yyyyyy";

$connection=mysql_connect("localhost", "root", "password") or die (mysql_error());

$db=mysql_select_db($db_name, $connection) or die (mysql_error());

foreach ($_POST[bid_status] as $key=>$v_bs){

$v_bs=implode(',',$_POST['bid_status']);

}

foreach ($_POST[equipment] as $key=>$v_e){

$v_e=implode(',',$_POST['equipment']);

}

echo "about to insert into database";

//line 62

$sql="INSERT INTO $table_name

(project_name, date, bid_status, engineer, owner, equipment, odor_value, pumps_value, service_value, project_value, comments) VALUES

('$_POST[project_name]','$_POST[date]','$v_bs','$_POST[engineer]', '$_POST[owner]', '$v_e','$v_odor', '$v_pumps', '$v_service', '$project_value', '$_POST[comments]')";

$result=mysql_query($sql, $connection) or die (mysql_error());

Ch0cu3r · March 5, 2014

Change $engineerList .= '<option value>'. to $engineerList .= '<option>'.

buchberg · March 5, 2014

Thank you so much. That did it. This is my first go at mysql fetch so this is very much appreciated. You are great.

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option value as part of form block

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