Problems with PHP's array_shift in an HTML script

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Problems with PHP's array_shift in an HTML script

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I have a problem that I can not solve and I hope you can help me:

I am working on Wordpress with PHP and HTML to extract data from a server with MYSQL and then to show this data on screen in a graphic thanks to Google charts.

Both taking data and storing it in an array works. Within php I make a short loop of 10 laps to show me the first 10 values of the array:

for ($i=0; $i<10; $i++){
   foreach ($elemento = array_shift($rIN) as $v1){
      echo "$v1--";
   }
}

This code works perfectly and shows me the first 10 values of the array as I need.

Now yes, the problem is when I need to do the same thing in an HTML script.

for(i=0,i<10,i++){
         data.addRow([i,<?php foreach ($elemento = array_shift($rIN) as $v1){echo "$v1";}?>]);
         }

In this case, google charts only graph me in all the values of x (0,1,2,3 .., 10) exactly the first value of the array only, that is to say, graph only a horizontal straight line.

If within the loop I add another data.addRow line exactly like the previous one, I graph the first and the second value in this way:

In summary, every time you start the loop again it is as if you had never applied the array_shift and the array was never modified.

In google charts (JS with PHP):

google.charts.load('current', {packages: ['corechart', 'line']});
google.charts.setOnLoadCallback(drawHour);
google.charts.setOnLoadCallback(drawDay);
function drawHour() {
  var data = new google.visualization.DataTable();
  data.addColumn('string', 'Hora');
  data.addColumn('number', 'IN');
  data.addColumn('number', 'OUT');
  for (i=0; i<1; i++){
    data.addRow([i,<'?php foreach ($elemento = array_pop($rIN) as $v1){echo "$v1";}?>,<'?php foreach ($out = array_pop($rOUT) as $v2){echo "$v2";}?>]);

html php google-chart-api

asked by Pablo Toledo 15.12.2017 в 16:44

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1 answer

Django filtering last JSONField object Doubt with Flex in header

score 0 · Answer 1

The problem lies in the order of execution of each language.

PHP is compiled on the server ( runs first ) and the result is returned to the browser ('agent making the solution').
The browser takes that result returned by the server and executes JS (among other things).

That is to say that, on each round of for of JS , the values printed by PHP are already in the document, and they will always be the same.

Solution:

You could generate the code JS with PHP .

Example:

var data = new google.visualization.DataTable();
data.addColumn('string', 'Hora');
data.addColumn('number', 'IN');
data.addColumn('number', 'OUT');
<?php
for ($i=0, $i<1, $i++) {

    echo "data.addRow([$i,";
    foreach ($elemento = array_shift($rIN) as $v1) { echo "$v1"; }
    echo ",";
    foreach ($out = array_pop($rOUT) as $v2){echo "$v2";}
    echo "]);";
} ?>