PHP function confusion -

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मेरे पास निम्न परिदृश्य (सरलीकृत) है: 

  फ़ंक्शन परिवर्तन फर्म ($ फलों) { changeAgain ($ फल); } फ़ंक्शन परिवर्तनआगेन ($ फलों) {$ फल = "ऑरेंज"; } मुख्य: $ फल = "ऐप्पल"; changeFruit ($ फल); $ फोकस गूंज / "एप्पल" के रूप में दिखाई देगा, मैं इसे "ऑरेंज" के रूप में कैसे दिखा सकता हूं ??

संपादित करें: जोड़ने के लिए हटाएं SCENARIO वापस स्टेटमेंट्स का उपयोग नहीं कर सकता - केवल & amp; $ चर

धन्यवाद! मैट म्यूएलर

जब आप कुछ पास करते हैं जो PHP में फ़ंक्शन के लिए ऑब्जेक्ट नहीं है, तो php एक प्रतिलिपि बनाता है फ़ंक्शन के भीतर इसका उपयोग करने के लिए।

इसे किसी प्रति का उपयोग नहीं करने के लिए, आपको एक PHP संदर्भित करने की आवश्यकता है।

यह & amp; ऑपरेटर

<पूर्व> फ़ंक्शन परिवर्तन फर्म (और $ फलों) {changeAgain ($ fruit); } फ़ंक्शन परिवर्तनआगेन (और $ फलों) {$ fruit = "ऑरेंज"; } $ फल = "ऐप्पल"; changeFruit ($ फल); प्रतिफल $ फलों;

फ़ंक्शन के रिटर्न वैल्यू का उपयोग करने के लिए यह अधिक समझदार और बेहतर अभ्यास होगा (क्योंकि इससे चीजों को पढ़ने में आसान होता है) 

  फ़ंक्शन परिवर्तन फर्म ( $ फलों) {वापसी परिवर्तनअगेन ($ फलों); } फ़ंक्शन परिवर्तनअगैन ($ फलों) {/ / $ $ फल = "ऑरेंज" यहां फल के साथ और अधिक रोचक कार्य करें; वापसी $ फलों; } $ फल = "ऐप्पल"; $ फल = परिवर्तनफुट ($ फलों); प्रतिफल $ फल

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