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PHP+MYSQL!? How better to output a couple of records?

ajax mysql records

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5 replies to this topic

#1 Stasonix


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Posted 22 November 2011 - 03:03 PM

for example, I have query like this

$sql = "SELECT `name`,`surname`,`age` FROM `table` WHERE (`name`='".$name,"') AND (`age`=20)";
$query = mysql_query($sql);
$result = mysql_fetch_array($query);

so I need to echo only 2 records is name and age and to recieve it in JQuery(ajax-method), what I am gonna do next?
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#2 John


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Posted 22 November 2011 - 08:28 PM

Iterate through the results echoing name and age.
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#3 codehutch


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Posted 22 November 2011 - 08:50 PM

while ($result = mysql_fetch_array($query, MYSQL_ASSOC)) {
    echo $result["name"] . " - " . $result["age"];
I don't know what you mean about the "ajax thing" though. Maybe you can make it more clearer.
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#4 Calgon


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Posted 23 November 2011 - 03:15 AM

Create a separate page which calculates everything using the code codehutch provided, and then use jquery's ajax functions to echo out the results dynamically, here's an example:

  type: "GET",
  url: "page_query_calc_thing.php",
}).done(function( msg ) {

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#5 SoN9ne


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Posted 23 November 2011 - 08:46 AM

If you are using it VIA AJAX, do not echo the data. echo the data as JSON. This is a standard for handling data VIA AJAX and it allows you more control over the data.
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"Life would be so much easier if we only had the source code."

#6 Orjan


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Posted 23 November 2011 - 03:18 PM

Agree with SoN9ne. You just read out your data and fill an array with it, and then, there are json_encode() function, just echo it's result to the ajax call.

---------- Post added at 12:18 AM ---------- Previous post was at 12:17 AM ----------

unnecesary post as I missed the SoN9nes link to the function, still I agree with him.
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Also tagged with one or more of these keywords: ajax, mysql, records

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