I have an array of arrays like below and I want to give them as arguments of array_intersect like below but this is not dynamic do you have better suggestions? Answer You can use call_user_func_array like this $intersect = call_user_func_array(‘array_intersect’,$array);
Tag: array-intersect
Compare 2 arrays containing rows of data and displays the matched records
I am trying to figure out a way that I could match the ‘userid’ on both arrays. Then the ones that match, keep the rows in MongoDBResults. Desired output: I tried: also Answer You can index on userid and then compute the intersection of the keys:
Why does array_uintersect() compare elements between array1 & array2, array1 & array1, and array2 & array2?
Test script Actual Result Expected Result In other words, what I am expecting to be passed to the callback is the current element of the left array, and the current element of the right array. Furthermore, I would expect the same logic to apply if I were to pass an additional array to array_uintersect – one more argument being passed
Compare two 2D arrays & get intersection and differences
I have two multidimensional arrays with me.. I want to split above given arrays into three different arrays.. Answer Convert arrays to a format, where array index is the sight_id: Calculate the differences and intersection: Convert arrays back to your format:
Array Intersection – only once
Newbie here, so I have a problem with the array_intersect function. I am comparing two arrays with intersection but I get some results twice. For example: 2 arrays $result = array_intersect($array1, $array2); This returns: But I want this: I want this to return apple and orange ONCE each. Currently it returns apple, then orange, and then another orange. Am I
array_uintersect() gives unexpected results when callback only returns 0 or 1
I have a custom callback in my array_uintersect() call because I need to case-sensitively compare strings in two elements while comparing rows between two multi-dimensional arrays. OUTPUT I can’t understand why this code doesn’t produce the correct output which should be an array with “B”, “C” and “D” elements, because only “A” element is different from $vieja to $nueva… if