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up vote 4 down vote favorite

I am using php database connectivity with "$row" as an array that stores name as a field in it .But after lots of searching i dint get how to pass php variable in angular js here's the code ,I am trying to achieve.how is it possible i am new to this your help will be appreciated.

<?php

$con = mysqli_connect('localhost','root','','practice');

if (!$con) {
    die("couldnt connect". mysqli_error);
}

$q= "select * from test";
$result= $con->query($q);

if ($result->num_rows>0) {

    while ($row = $result->fetch_assoc()) {

        echo $row['name']."<br>";

    }
}

print_r($result);

echo json_encode($result);

?>
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.0.4/angular.js">
</head>
<body>

<ul ng-init="names = <?php echo htmlspecialchars(json_encode($row)); ?>">
<li ng-repeat="x in names">
   <p>{{x.name}}</p>
</li>
</ul>
</body>
</html>
share|improve this question
    
Your script needs some serous tidy up, your are echoing/printing lots of stuff before even your html starts. if there are purely to test your DB outcome then why to mix them the with <br> tags. anyway what is the out put of echo $row['name']."<br>";? when you check the page source will you see any data from the db? change that line to print_r($row['name']); and let me know if the data is fetched and printed (on your html page source) then we will get into solving the problem :) – Ali Jul 5 '15 at 8:54
1  
no @ali it is all fine with database output $row['name'] is perfectly fetching the data but problem is with js how can i show that array data of $row['name'] in angular js – Share fun Jul 5 '15 at 9:01
    
ok cool, next thing will be including the angular js script, something like <script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.0.4/an‌​gular.js"> – Ali Jul 5 '15 at 9:07
    
yeah my mistake SORRY but still its showing this "[]" only as a output @Ali – Share fun Jul 5 '15 at 9:15
    
whats the value of <?php echo htmlspecialchars(json_encode($row)); ?> ? – Ali Jul 5 '15 at 10:09
up vote 3 down vote accepted

Your php code needs a bit of improvement use this instead of that:

$arr = array();
if ($result->num_rows>0) {

while ($row = $result->fetch_assoc()) {

$arr[] = $row['name'] ;

}
}

Then echo it as

<ul ng-init="names = <?php echo htmlspecialchars(json_encode($arr)); ?>">
share|improve this answer
    
Yeah that was quiet easy . Thanks for pointing out the error it helped alot thanks – Share fun Jul 16 '15 at 17:27
up vote 2 down vote

As discussed earlier, you need to format your data in the names's to;

  1. be a valid valid JSON
  2. provide a field name since you are seeking it in your loop (ng-repeat)

So at the moment the value assigned to your "names" is "eden frank matt" as you mentioned in your comment. this needs to be changed to something like:

[
    {name: 'eden'},
    {frank: 'eden'},
    {matt: 'eden'}
]

so when you check the html source we should have a collection of lets say "people" and then going through all of its element and point to their field "name" :

<!DOCTYPE html>
<html>
<head>
    <script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.0.4/angular.js">
    </script>
</head>


<body ng-app>

<ul ng-init="people = [{name:'eden'}, {name:'frank'}, {name:'matt'}]">
    <li ng-repeat="person in people">
        <p>{{person.name}}</p>
    </li>
</ul>

</body>
</html>
share|improve this answer
    
I know this that's the basic thing but if i get a lot of database in an array with lots of fields in it then php database connection is required and then convert it into angular js array will show up that data. I want to achieve php with angular js . i know i can do this in php alone . but its mandatory in my project – Share fun Jul 5 '15 at 11:20
    
i didn't suggest to write the JSON manually @Sharefun. you have to pars your result from the database in a way to be a vlid JSON with the fields your going to look for them further on. the example in my answer is when you are looking into the HTML source in the browser and not your code – Ali Jul 5 '15 at 11:28
    
i've asked you what you see when you check the html source (in the browser i meant) . so again, when you open your page in a web browser, when you view the source (in the web browser) what will you see instead of <ul ng-init="names = <?php echo htmlspecialchars(json_encode($row)); ?>"> ? it should be <ul ng-init="names = ....."> .which the (.....) is the bit whoch I'm interested to know :) – Ali Jul 5 '15 at 11:33
    
Okay @Ali that's fine . but their is another way i can choose may be like using $http.get() and success function to this i dont know how use i am new to angular js and cant find any exact solution for this . If you know then please tell me. – Share fun Jul 5 '15 at 11:37
    
okay it shows "null" it means nothing is coming up in $row but php is showing the data but as i pass this to angular js it shows null . I MUST HAVE DOING SOMETHING WRONG . cant figure it out. – Share fun Jul 5 '15 at 11:41
up vote 0 down vote

I had similar requirement, and the below worked for me.

data-ng-click="editField(product, '<?php echo $_SESSION["login_user"] ?>')"
share|improve this answer

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