This is php script that fetches a table values from mysql (one row). & echoes it as JSON
<?php
$username = "user";
$password = "********";
$hostname = "localhost";
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to
connect to MySQL");
$selected = mysql_select_db("spec",$dbh) or die("Could not select first_test");
$query = "SELECT * FROM user_spec";
$result=mysql_query($query);
$outArray = array();
if ($result) {
while ($row = mysql_fetch_assoc($result)) $outArray[] = $row;
}
echo json_encode($outArray);
?>
this is HTML file to receive & print json data.
src=".4.2/jquery.min.js">
//$('document').ready(function() {
function Preload() {
$.getJSON("http://localhost/conn_mysql.php", function(jsonData){
$.each(jsonData, function(i,j)
{ alert(j.options);});
});}
// });
</script></head>
<body onLoad="Preload()">
</body>
</html> >
This is php script that fetches a table values from mysql (one row). & echoes it as JSON
<?php
$username = "user";
$password = "********";
$hostname = "localhost";
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to
connect to MySQL");
$selected = mysql_select_db("spec",$dbh) or die("Could not select first_test");
$query = "SELECT * FROM user_spec";
$result=mysql_query($query);
$outArray = array();
if ($result) {
while ($row = mysql_fetch_assoc($result)) $outArray[] = $row;
}
echo json_encode($outArray);
?>
this is HTML file to receive & print json data.
src="http://ajax.googleapis./ajax/libs/jquery/1.4.2/jquery.min.js">
//$('document').ready(function() {
function Preload() {
$.getJSON("http://localhost/conn_mysql.php", function(jsonData){
$.each(jsonData, function(i,j)
{ alert(j.options);});
});}
// });
</script></head>
<body onLoad="Preload()">
</body>
</html> >
- did you ask this already? stackoverflow./questions/3963057/… – Phill Pafford Commented Oct 19, 2010 at 19:48
- no, code is same about which I ask other early things but now with some clarity & addition in code, I have new question that is ahead from previous ones. But if there is something special to care about it, kindly let me know. – XCeptable Commented Oct 19, 2010 at 19:58
3 Answers
Reset to default 8Your PHP needs to actually put all the rows together:
$query = "SELECT * FROM user_spec";
$result=mysql_query($query);
$outArray = array();
if ($result) {
while ($row = mysql_fetch_assoc($result)) $outArray[] = $row;
}
echo json_encode($outArray);
Your Javascript needs to look at each of the rows..
$.getJSON("/whatever.php", function(jsonData) {
for (var x = 0; x < jsonData.length; x++) {
alert(jsonData[x].options);
}
});
mysql_fetch_assoc will only return a single row from the database. You will need a loop to retrieve all rows:
$data = array();
while ($row = mysql_fetch_assoc($result)) {
// add some or all of $row to the $data array
}
echo json_encode($data);
<?php
$username = "user";
$password = "********";
$hostname = "localhost";
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to connect
to MySQL");
$selected = mysql_select_db("spec",$dbh) or die("Could not select first_test");
$query = "SELECT * FROM user_spec";
$result=mysql_query($query);
$_ResultSet = array();
while ($row = mysql_fetch_assoc($result)) {
$_ResultSet[] = $row;
}
echo json_encode($_ResultSet);
?>
and your jQuery would looks like :
$.getJSON("/yourscript.php", function(data) {
$.each(data, function(i, j) {
// use: j.columnName
});
});
Good luck
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